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owlofcreamcheese
12-04-2002, 06:12 PM
this actually isn't even a question more of a "huh, thats weird, hes absolutely right" moment

a while back I took a class about infinity.

say god got bored and made a bunch of chips with numbers written on them 1 to infinity (he can do that... hes god... he created an infinite amount of chips... )

and yes you can do an infinite amount of things in a finite amount of time... xeno's paradox and all. (just take an hour on the first one, half an hour on the second one, 1/4th an hour on the 3rd one... as the number you do goes to infinity the time it takes becomes exactly 2 hours)

so god takes his chips... and makes a big dish for them... and puts them in a pile next to them....

he plans to sort them...

his plan is to take two out of his pile and put them both in the dish... then look in the dish and take out the biggest one.

so he does that for every chip (takeing two hours) and he takes out the biggest one

turn one puts in 1 and 2... takes out 2....
second turn puts in 3 and 4 takes out 4
third turn puts in 5 and 6 takes out 6

ect ect ect...

when hes done... he looks at his dish and says "wow, I have every single odd number in my dish" (since 1,3,5, ect never were the highest in the dish)

now jesus comes in and says "dad! thats not fair, takeing out the biggest one... its unfair"

so god comes up with a new way to sort them for the next time he plays

he takes out the smallest number each time.

so he puts in 1 and 2 and takes out 1 (so there is one in the dish)
next he puts in 3 and 4 and takes out 2 (two in the dish now)
next 5 and 6 and takes out 3 (3 chips left in the dish)
7 and 8 takes out 4 (4 left in the dish)

while hes doing this... being all smart he says "every time I do it, there is one more chip left in the dish than last time" being even more smart he knows that after doing that an infinite number of times... there will be an infinite number of chips in it...

so he calls over jesus "hey jesus! look at this! there will be an infinite number of chips in this dish when I am done!"

jesus says "oh yeah, name one?"

god says... well I bet 1 billion is in there....
jesus points out... "nope you took that out the billion and first turn same for a trillion or a jillion"

question is... how does that work? its odviously correct, but my brain isn't quite sure what that means....
I mean the whole infinity fractions thing is the base of inergal calculus so thats right...

derr... the solution is that no chips remain in the dish after its done... but derr..... derr... my head hurts...

Tyrrell McAllister
12-04-2002, 06:30 PM
Yep. Infinity's a weird number. Chips are put in twice as fast as they're taken out, so the number of chips in the dish is always increasing; but every chip is eventually taken out, so in the end none are left. If we were watching God do this over the course of a minute by doing the first swap at 1/2 minutes, the second at (1/2 + 1/4) minutes, the third at (1/2 + 1/4 + 1/8) minutes, etc, the rate at which the number of chips was increasing would appear to accelerate exponentially, until suddenly, at t = 1 minute, they'd all disappear.

loupdebois
12-04-2002, 06:58 PM
Here's my uneducated take on it (great spatial skills, no time for class):

Let's not conceive of infinity. You're not supposed to actually picture it, I'm told one can get brain cancer from doing so. Infinity is beyond conception. However, what I noted about your pattern is that you're basically keeping the numbers going up by one, while the range of integers in your chip bowl is always an uninterrupted sequence between x and 2x (x being the chip that gets taken out "next")... since infinity is inconceivably huge, no number that God could name could be half of infinity (therefore waiting to be taken out rather than already out). Clever Tyrell. No chips in the bowl... at least none that would fit inside our universe. Infinitely big chips. Nifty.
Personally, I think if he did it in a minute, they wouldn't all disappear, because the bowl is overflowing with half of an infinite number of chips... and even silly people know that half of infinity is still infinity. SO.... the universe explodes or something. oops.

ultrafilter
12-04-2002, 07:00 PM
Yep. It's kinda weird, ain't it?

Urban Ranger
12-04-2002, 08:53 PM
You can't.

Since it is infinity, there is no biggest number.

Urban Ranger
12-04-2002, 08:58 PM
Originally posted by owlofcreamcheese
and yes you can do an infinite amount of things in a finite amount of time... xeno's paradox and all. (just take an hour on the first one, half an hour on the second one, 1/4th an hour on the 3rd one... as the number you do goes to infinity the time it takes becomes exactly 2 hours)

And you can't infinite amount of things in a finite amount of time. For the last thing, you have to do it in 1/infinity, which bombs out.

David Simmons
12-04-2002, 09:13 PM
Well you know, infinity is not a number so it isn't surprising that the rules of arithmetic, which deals with numbers, don't apply.

t-keela
12-04-2002, 09:33 PM
I think this is explained in
http://www.mathpages.com/rr/s3-07/3-07.html
http://mathforum.org/isaac/problems/zeno1.html
one of these links explains how an infinite number of events can be done in a finite amount of time.
Granted, it's just theory and based on ancient physics, but to say it's not Xeno's paradox...well.

yes, you can>no, you can't >yeah>no>...;)

I think considering the lack of technology and even a basic understanding of quantum mechanics he did allright.

knock knock
12-04-2002, 10:29 PM
Originally posted by owlofcreamcheese
while hes doing this... being all smart he says "every time I do it, there is one more chip left in the dish than last time" being even more smart he knows that after doing that an infinite number of times... there will be an infinite number of chips in it...

so he calls over jesus "hey jesus! look at this! there will be an infinite number of chips in this dish when I am done!"

jesus says "oh yeah, name one?"

god says... well I bet 1 billion is in there....
jesus points out... "nope you took that out the billion and first turn same for a trillion or a jillion"

question is... how does that work? its odviously correct, but my brain isn't quite sure what that means....
Maybe that's how God created the supernatural numbers.

The what? I'd provide a link, but I can't find any good sites out there. Basically, supernatural numbers can be thought of as "infinitely big numbers." Real numbers can be expressed in the form xx.xxxx.... (e.g. pi = 3.141592....), and supernatural numbers can be expressed in the form ....xxxx.xx (you can't write them as a numeral). You could think of infinitesimals like dx as the inverses of supernatural numbers. The best discussion I've seen of supernatural numbers (and a great book that looks at all sorts of questions like this) is Gödel, Escher, Bach: An Eternal Golden Braid (http://www.amazon.com/exec/obidos/tg/detail/-/0465026567/ref=lib_rd_btb/103-1763070-1462233?v=glance&s=books) by Douglas Hofstadter.

God does have to have an "infinite number" of chips in the dish, as is obvious (more or less) if you ignore the numbers on the chips and just count the number of chips. A sorta rigorous way of thinking about this is that, if you name any number n, then I can find some time before the 2 hr. limit when there are more than n chips in the dish, and, since the # of chips in the dish is always increasing (or at least non-decreasing), there will be more than n chips in the dish at any later time. Since the # of chips grows greater than any number n as you approach the 2-hr mark, it must be approaching infinity.

owlofcreamcheese
12-04-2002, 11:31 PM
urban ranger... well its related to xeno.... alot of calculus only works because you can add an infinite number of things and get a whole number.

ultrafilter
12-04-2002, 11:37 PM
Originally posted by owlofcreamcheese
urban ranger... well its related to xeno.... alot of calculus only works because you can add an infinite number of things and get a whole number.

No, no, a thousand times no! Calculus works because as you add more and more terms, the sum gets *closer* to a specific value. There is no adding of an infinite number of terms.

Sorry if I sound harsh, but this is probably one of the most confusing facts about math, and I've seen a few heated arguments arise from it.

t-keela
12-04-2002, 11:43 PM
I think the trick is in the definition of infinity in different contexts. It can be used as either countable infinity or not countable. By limiting it to a set (ie:a space of time with a beginning and an end) you have changed the context to where it can be counted. This way you can actually "measure" infinity in a finite setting.

So, which set is largest...the (infinite) set of even integers or the I-set of odd integers or the set that contains all integers?

Tyrrell McAllister
12-05-2002, 03:14 AM
God does have to have an "infinite number" of chips in the dish, as is obvious (more or less) if you ignore the numbers on the chips and just count the number of chips. A sorta rigorous way of thinking about this is that, if you name any number n, then I can find some time before the 2 hr. limit when there are more than n chips in the dish, and, since the # of chips in the dish is always increasing (or at least non-decreasing), there will be more than n chips in the dish at any later time. Since the # of chips grows greater than any number n as you approach the 2-hr mark, it must be approaching infinity.

I don't think that this is quite right. It is one thing to say that,
For every number n, there is a time t such that more than n chips are in the dish at time t. This is true. But to say that there are ever infinitely many chips in the dish (ie, that "God does have to have an "infinite number" of chips in the dish") is to say that
There is a time t such that, for every number n, more than n chips are in the dish at time t. This is not true.

mrcrow
12-05-2002, 03:25 AM
Originally posted by Urban Ranger
You can't.

Since it is infinity, there is no biggest number.
or small
infinity is not a dimension
a concept to defy dimensionally related concepts:)

Achernar
12-05-2002, 06:07 AM
Originally posted by loupdebois
Let's not conceive of infinity. You're not supposed to actually picture it, I'm told one can get brain cancer from doing soUh-oh. A lot of us are in trouble, then.

ultrafilter
12-05-2002, 08:27 AM
Originally posted by t-keela
I think the trick is in the definition of infinity in different contexts. It can be used as either countable infinity or not countable. By limiting it to a set (ie:a space of time with a beginning and an end) you have changed the context to where it can be counted. This way you can actually "measure" infinity in a finite setting.

So, which set is largest...the (infinite) set of even integers or the I-set of odd integers or the set that contains all integers?

They're all the same size. The problem deals only with countable infinities.

knock knock
12-05-2002, 08:53 AM
Originally posted by Tyrrell McAllister
I don't think that this is quite right. It is one thing to say that,
For every number n, there is a time t such that more than n chips are in the dish at time t. This is true. But to say that there are ever infinitely many chips in the dish (ie, that "God does have to have an "infinite number" of chips in the dish") is to say that
There is a time t such that, for every number n, more than n chips are in the dish at time t. This is not true. During the time period when 0 (< or =) t < 2 hr., you're right, there are never "an infinite number of chips" in the dish. However, as the time approaches the 2-hr mark, the number of chips in the dish approaches infinity (i.e. increases without bound). And at the time t=2 hours, how many chips would you say are in the dish? I would say that there are infinitely many, since, for any n, there are more than n chips in the dish at t=2 hr.

moes lotion
12-05-2002, 10:36 AM
The OP's second example is cute, but it seems to me that there is a problem in trying to equate the cardinality of an infinite set {1, 2, 3,...} with the sum of an infinite series 1/2, 1/4, 1/8, .... Like the Energizer Bunny, the infinite set just keeps on going. When you say that the sum of the series exists, you mean that the total can be made arbitrarily close to a value by including enough terms. The problem with trying to equate this to how many disks are left in the bowl is that there is no last one that gets put in the bowl - the number is always increasing.

As t-keela mentions, there are different "sizes" of infinity. The first infinity is called countable because a countably infinite set is one which can be brought into a 1-to-1 corespondence with the set of positive integers {1, 2, 3, ...} Thus the set of all +ve odd integers, the set of all +ve even integers and the set of all +ve integers are each countably infinite, and thus the same "size". Any odd number can be expressed as 2n - 1, any even number as 2n for n = 1, 2, 3, ...
thus there is a 1-to-1 corespondence between the three sets.

IIRC correctly, the next "size" of infinity is represented by the set of real numbers. It is possible to show that one cannot establish a 1-to-1 corespondence between the set of real numbers x such that 0 < x < 1 and the set of +ve integers, thus there are uncountably many real numbers in this interval. Apparently there are even higher kinds of infinity, but at that point my brain was full so I asked to be excused from class.

DrMatrix
12-05-2002, 10:56 AM
Originally posted by moes lotion

IIRC correctly, the next "size" of infinity is represented by the set of real numbers. Well . . . The cardinality of the reals is the next cardinal number, if you accept Cantor' Continuum Hypothesis.The Continuum Hypothesis is independent of the other axioms of set theory. Since it is independent it can be assumed true or assumed false. Neither will result in a contradiction (that wasn't already there). The set of reals is strictly larger than the set of integers with or without CH. If you assume CH is false, then you are assuming there are sets of reals that is uncountable, but still smaller than the set of reals.

For any set A, the power set of A (the set of all subsets of A) is larger than A. The Generalized Continuum Hypothesis says that if A is infinite, the next larger "size" set is the power set of A. GCH, like CH is independent of the other axioms. It can be assumed true or assumed false.

RM Mentock
12-05-2002, 11:06 AM
Originally posted by owlofcreamcheese
a while back I took a class about infinity.
Huh? You took a class in it, and you have to ask this question? Didn't they teach you everything?

KidCharlemagne
12-05-2002, 11:19 AM
Aren't different size infinities called alephs or something?

Also doesn't this puzzle make a jump in considering the universe to be continuous up until time zero when the assumption becomes that its discrete?

aahala
12-05-2002, 11:34 AM
Originally posted by RM Mentock
Huh? You took a class in it, and you have to ask this question? Didn't they teach you everything?

The reason the OP couldn't answer his own question, was that he missed the class lab sessions where the infinity experiment is being done.

Never fear, even though he missed all the labs so far, he hasn't missed even .00000001% of them.

KidCharlemagne
12-05-2002, 11:38 AM
Originally posted by aahala
The reason the OP couldn't answer his own question, was that he missed the class lab sessions where the infinity experiment is being done.

Never fear, even though he missed all the labs so far, he hasn't missed even .00000001% of them.

Nope, I think the day before was the Xeno's Paradox lecture and his professor told him to try the old "keep coming half way to class" trick. I believe he's still realllllly cloooose to the classroom.

Tyrrell McAllister
12-05-2002, 01:42 PM
Originally posted by knock knock

And at the time t=2 hours, how many chips would you say are in the dish? I would say that there are infinitely many, since, for any n, there are more than n chips in the dish at t=2 hr.
I say that there are zero chips in the dish, because, given any particular chip, that chip was removed at some time before the 2 hour mark (to use your coordinates). Hence, no chip can still be in the bowl by the time 2 hours rolls around. Therefore, at precisely 2 hours, the bowl is completely empty. In particular, it never has infinitely many chips in it.

erislover
12-05-2002, 01:43 PM
No, no, a thousand times no! Calculus works because as you add more and more terms, the sum gets *closer* to a specific value. There is no adding of an infinite number of terms.

Sorry if I sound harsh, but this is probably one of the most confusing facts about math, and I've seen a few heated arguments arise from it.A fun discussion, IMO:
http://www.friesian.com/calculus.htm

ultrafilter
12-05-2002, 01:55 PM
Originally posted by erislover
A fun discussion, IMO:
http://www.friesian.com/calculus.htm

From that page
dy/dx therefore represents zero/zero, which is ordinarily a meaningless relationship in mathematics: zero divided by anything is zero; and anything divided by zero is often said to be "undefined," which is a polite (or wimpy) way of saying "infinite."

I got as far as that. When I stopped grimacing, I closed the window.

Tyrrell McAllister
12-05-2002, 03:19 PM
You should give that page another chance, ultrafilter. That paragraph is just expressing the apparent difficulty with calculus that mathematicians struggled with before the limit definitions. It then goes on to discuss how, in the 20th century, infinitesimals were given a respectible footing, and outlines a particular technique for doing this. The so-called non-standard analysis (http://mathworld.wolfram.com/NonstandardAnalysis.html) that studies these objects does actually have applications in other fields.

Liberal
12-05-2002, 03:20 PM
Ultrafilter quoted from Friesian.com:

...and anything divided by zero is often said to be "undefined," which is a polite (or wimpy) way of saying "infinite."I don't buy that.

12/3 = 4 can be understood to mean that 3 can be subtracted from 12 4 times before exhausting the value of 12. But how many times for 12/0 can 0 be subtracted from 12 before exhausting the value of 12? Not even an "infinite number" (if there were such a thing) of times will do.

-----

Nice puzzle, by the way, Owlofcreamcheese. Is it original?

ultrafilter
12-05-2002, 03:36 PM
Originally posted by Tyrrell McAllister
You should give that page another chance, ultrafilter. That paragraph is just expressing the apparent difficulty with calculus that mathematicians struggled with before the limit definitions. It then goes on to discuss how, in the 20th century, infinitesimals were given a respectible footing, and outlines a particular technique for doing this. The so-called non-standard analysis (http://mathworld.wolfram.com/NonstandardAnalysis.html) that studies these objects does actually have applications in other fields.

I actually did go back and read the whole thing. My first impression was that the author was advocating the position she described, but apparently I was wrong.

Lib: That's the reason I was grimacing. "undefined" doesn't mean "infinite", it means "undefined".

photopat
12-05-2002, 04:18 PM
Even assuming an infinite number of objects - chips- are being placed in a bowl...

If each one is placed in half the time of the one before, and the first one is in an hour, the eleventh will be given 3.5 seconds, the fourteenth gets .44 seconds, etc., until as the 2 hour mark nears, billions will be placed every second, but because they are infinite, there will never be an end and 2 hours will never be reached. It will be approached and to human perception will be reached, but...not quite, and never, and the chips will never all be in the bowl, because there will always be more. That's the thing about infinity. There's always more.

The way I was taught about division by zero was by dividing 1 by progressively smaller fractions. 1/.1 = 10, 1/.01 = 100, 1/.001 = 1000, and so on. The closer the denominator gets to zero, the greater the product, so that it approaches infinity, but never reaches it. So, any number divided by zero is effectively = infinity.

ZenBeam
12-05-2002, 04:31 PM
The way I was taught about division by zero was by dividing 1 by progressively smaller fractions. 1/.1 = 10, 1/.01 = 100, 1/.001 = 1000, and so on. The closer the denominator gets to zero, the greater the product, so that it approaches infinity, but never reaches it. So, any number divided by zero is effectively = infinity.What about the sequence 1 / -0.1 = -10, 1 / -0.01 = -100, 1 / -0.001 = -1000, ....? Doesn't this imply 1 / 0 is negative infinity?

ultrafilter
12-05-2002, 04:33 PM
In order that we agree that 1/0 is infinity in the limit, it would need to be the case that f(x)/g(x) goes to infinity whenever there is a c such that, as x goes to c, f(x) goes to 1 and g(x) goes to zero.

But that isn't the case. Consider f(x) = cos(x) and g(x) = x. For x > 0, f(x)/g(x) > 0, but for x < 0, cos(x)/x < 0. So if this function has a limit as x -> 0, it must be zero. But that's not the case--in fact, |cos(x)/x| goes to infinity as x -> 0. So cos(x)/x goes to -infinity as we approach from the left, and infinity as we approach from the right. No limit.

ultrafilter
12-05-2002, 04:36 PM
btw, my reasoning is that if, as x -> c, f(x) -> a and g(x) -> b (b != 0), then f(x)/g(x) -> a/b as x -> c.

photopat
12-05-2002, 05:01 PM
See, this is why I should never get involved in math questions.

erislover
12-05-2002, 05:03 PM
It is possible, in certain circumstances, to evaluate limits of the form f(x)/g(x) where f and g go to zero at the limit in question (or infinity over zero, or zero over infinity, or blah blah blah). L'Hopital's rule, right? Right. So. Are there any functions for which, no matter how many times one applies the rule (since it allows for recursion), one always comes up with 0/0 (or whatever)? I understand it can fail, but the only times I've seen the rule unable to be applied was when a derivative no longer yielded a continuous function.

Total hijack there, sorry, but that rule fascinates me.

Hari Seldon
12-05-2002, 05:15 PM
Ah there is so much confusion over infinity. First place it is not a number, so talking about 1/0 = infinity is nonesense. That said, it it is certainly possible that some limits exist, such as f(x)/g(x) as x approaches some place that both f(x) and g(x) are 0 (or even if they both approach--not reach--infinity). Second place, there is no problem in positing and working with infinite and infinitesimal numbers. And 1/infinitesimal will be an infinite number and all the rules of arithmetic will be satisfied. But you still cannot divide by 0. All of which has nothing to do with the OP, which is one of the apparent paradoxes of the infinitel

ultrafilter
12-05-2002, 05:18 PM
Originally posted by erislover
It is possible, in certain circumstances, to evaluate limits of the form f(x)/g(x) where f and g go to zero at the limit in question (or infinity over zero, or zero over infinity, or blah blah blah). L'Hopital's rule, right? Right. So. Are there any functions for which, no matter how many times one applies the rule (since it allows for recursion), one always comes up with 0/0 (or whatever)? I understand it can fail, but the only times I've seen the rule unable to be applied was when a derivative no longer yielded a continuous function.

Total hijack there, sorry, but that rule fascinates me.

If you have two exponential functions, L'Hopital's rule won't help. But there are other means for that case.

knock knock
12-05-2002, 05:33 PM
so he calls over jesus "hey jesus! look at this! there will be an infinite number of chips in this dish when I am done!"

jesus says "oh yeah, name one?"

god says... well I bet 1 billion is in there....
jesus points out... "nope you took that out the billion and first turn same for a trillion or a jillion"The "god" that you speak of cannot be the Judeo-Christian God, for He is infallible, while this god erroneously says "well I bet 1 billion is in there." So he must be a pagan god. Nothing that your jesus says is erroneous, so it seems that he might be the true Jesus Christ. However, he is speaking with this pagan god, which is clearly forbidden by the commandments (the Bible is pretty clear that there is only one true god). So he must be a pagan jesus.

I say that there are zero chips in the dish, because, given any particular chip, that chip was removed at some time before the 2 hour mark (to use your coordinates). Hence, no chip can still be in the bowl by the time 2 hours rolls around. Therefore, at precisely 2 hours, the bowl is completely empty. In particular, it never has infinitely many chips in it.You agree that there are infinitely many chips in the dish in the first example, right? All the odd numbers = infinitely many chips. Imagine that I was watching him put chips in the bowl, and you were watching him take them out. I would note "he put two in...two more in...two in...2...2...2..." and pretty soon I'd be talking kinda fast. You would note "he took one out...one more out...one out...1...1...1..." and pretty soon you'd be talking fast. We would be giving all of the information relevant to the number of chips currently in the dish, since that number is completely determined by 1) the # of chips in the dish to begin with (zero), 2) the number of chips that go in the dish (and maybe when they go in), and 3) the number of chips that come out (and maybe when they come out). But we'd be saying the exact same thing in the first case, when all the odd numbered chips stayed in the dish, as in the second case, when I-don't-know-what chips end up in the dish. Since all relevant information is identical, the # of chips in the dish must be identical, i.e., infinite.

Here's a related problem. Say you have a sequence of functions. The first function f(1) is equal to one on the interval from 1 to 2, including 2 but not 1. That is, it equals 1 for x such that 1 < x (< or =) 2. It is 0 everywhere else. I'll write this as f(1) = 1 on (1,2] and 0 elsewhere. The second function is f(2) = 1 on (2,4] (and equals 0 elsewhere), the third f(3) = 1 on (3,6] (and is 0 elsewhere), the 647th f(647) = 1 on (647,1294] (and 0 elsewhere), and the nth is f(n) = 1 on (n,2n] (and f(n) = 0 elsewhere).

Note that the area under the nth function is equal to the number of chips in the dish after God's nth "move" of putting chips in & taking them out, and that the numbers on the chips in the dish after the nth move are the integers in the interval where f(n) equals 1. As n approaches infinity (increases without bound), the area under the function (equivalent to the number of chips in the dish) approaches infinity (since it always equals n). Now, you can't name any number where the function is nonzero (maybe you could say that it's nonzero on the interval from infinity to twofinity, but that would be nonsense), but the area does not suddenly become 0 (i.e. no chips in the dish).

We're looking at this problem in two different ways - I (like god) am looking at the number of chips in the dish (or the area under the function) for finite n, and taking the limit of that as the number of moves approaches infinity. You (like jesus) are looking at the particular chips in the dish (or the nth function) and taking a limit of that as n goes to infinity, and then looking at the area under that. In that case, the function approaches the function f that is equal to 0 everywhere, so the area under it is equal to zero. But when you switch between these perspectives, you're interchanging limits, and you aren't always allowed to do that without changing the answer. This is one of the cases where you can't do it.

So, in a way, we're both right. But, though it may be ok to just say that the two different limits have different values when you're looking at functions, it's harder to say that in the case of a concrete situation like this. You have to wonder, in this case, what interpretation fits the problem posed (i.e. which limit to take first). Since we're only looking at the number of chips in the dish, I see no reason to go through the intermediary of the numbers on the chips - just count. And when you count, God has infinitely many chips in the dish after 2 hours.

Cabbage
12-05-2002, 08:47 PM
And when you count, God has infinitely many chips in the dish after 2 hours.
Name one.

Liberal
12-06-2002, 04:51 AM
:D Cabbage, you are simply the best!

mrcrow
12-06-2002, 05:13 AM
the word infinite exists to describe what it can't
not finite..? jillion or whatever:D
abstract is the closest you will get...and then you can half that distance forever (is that a term) and never get there...which cant be because it would be finite.:cool:

Mangetout
12-06-2002, 05:15 AM
No discussion of infinity is complete without a reference to Hilbert's Hotel (http://www.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel) - a problem that is rather similar to the one outlined by the OP.

Mangetout
12-06-2002, 05:16 AM

mrcrow
12-06-2002, 05:18 AM
infinitely so:D

mrcrow
12-06-2002, 05:20 AM
??? no it didnt:(

mrcrow
12-06-2002, 05:21 AM
brkt missing?:eek:

mrcrow
12-06-2002, 05:23 AM

Mangetout
12-06-2002, 05:25 AM
I've started an ATMB thread about this problem; we needn't fill this one up with an inifinte number of posts on the topic.

mrcrow
12-06-2002, 05:27 AM

Achernar
12-06-2002, 05:40 AM

mrcrow
12-06-2002, 07:45 AM

me again
12-06-2002, 12:04 PM
Can't remember the book, but I remember infinity from it:

"Far to the north there is a big black rock, one hundred miles high and one hundred miles wide. Every thousand years a little bird flies to the rock to sharpen its beak.

"When all the rock has worn away only one second in a minute of infinity will have passed."

mrcrow
12-06-2002, 01:48 PM
Originally posted by me again
Can't remember the book, but I remember infinity from it:

"Far to the north there is a big black rock, one hundred miles high and one hundred miles wide. Every thousand years a little bird flies to the rock to sharpen its beak.

"When all the rock has worn away only one second in a minute of infinity will have passed."
eternity would be more appropriate
still there is no time measure for eternal things.:cool:

t-keela
12-07-2002, 09:01 AM
http://www.everything2.org/index.pl?node_id=800558

There's a link to the little bird parable.

"You've reached the end of the internet..." hmmm

Chronos
12-08-2002, 02:00 PM
"When all the rock has worn away only one second in a minute of infinity will have passed."No, no, no. It'll only take a finite time for the rock to be worn away. And even if that's only a second in a minute of some other time (whatever that means), we're still finite, here.

An illustration is in order, here. Consider Graham's Number, a very large number which has come up in theoretical mathematics. We'll need to introduce some new notation here, called arrow notation (I can't make arrows here, so I'll use carats instead). For starters, let 3^3 = 33. Then, let 3^^3 mean 3(33), 3^^^3 = 3(3(33)), etc. OK, now, take 3^3. Take that number, and put that many arrows in the next step, 3^^^^^^^^^^^^^^^^^^^^^^^^^^^3. Now take the number that results from that, and put that in the next step. Continue this for a total of 64 steps, and you've got Graham's Number.

If your brain hasn't yet exploded from the incredible hugeness of this number, you can now ask yourself this: How does this number compare to anything infinite? It's not "almost infinity", it's not even a tiny fraction of infinity. If you take away that many integers, the number of integers left over is not just "not diminished significantly", it's not diminished at all.

ZenBeam
12-09-2002, 04:54 PM
Cabbage wrote:And when you count, God has infinitely many chips in the dish after 2 hours.Name one.The"zero chips left" argument seems to depend on not being able to name a chip left. If you assume the continuum hypothesis is false, there is an order of infinity between the integers and the reals. So there must be a subset of the reals which has more elements than the set of integers, but fewer than the set of reals. In fact, there must be an infinite number of them. Yet it is apparently (http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=131395&perpage=40&pagenumber=2) impossible to specify such a subset (bolding mine, quoted post is two posts before my first post):
If CH is false, the cardinality of the reals is greater than aleph-1, and (by the axiom of choice) we can get a subset of the reals with cardinality aleph-1, but we can't construct such a subset.So just because you can't "name" something apparently doesn't mean it doesn't exist. Can you reconcile these two arguments?

Tyrrell McAllister
12-09-2002, 05:39 PM
But the deal with this set of chips in the bowl after two hours is that we proved, not only that it is impossible to name any chip in there, but moreover that each and every chip is demonstrably not in there. By "demonstrably", I mean that we can name the exact time at which any given chip is removed from the dish, and in each case, without exception, the time is before two hours have passed.

This is different from a counterexample set to CH. We may not be able to name any of the particular elements in such a set. But that is a far cry from being able to show that, given any real number, that real number is not in there. For if we could do such a thing, we would have shown that the set was in fact empty, and therefore not a counterexample to CH.

Cabbage
12-09-2002, 05:45 PM
ZenBeam, that's a good point; you're right that my "Name one" "argument", by itself, doesn't demonstrate that the box of chips is empty, for the reasons you mention.

However, I can demonstrate that those reasons don't apply in this situation. First of all, God only puts integers in the box, so we know that after two hours (that's the timespan we've been using, isn't it?) the chips left in the box are a subset of the integers. Any integer can be described (just write it out), and for any integer that can be describe, I can describe the time it was taken out of the box. Therefore, no integers are in the box, and, since only integers are in the box, the box must be empty.

knock-knock's flaw in his/her logic is using limits. If f(t) is the number of chips in the box, it's incorrect to say that the number of chips in the box after two hours is equal to lim f(t) as t goes to two (from the left). In general, you can't evaluate a function at a particular point by taking a limit at that point--that assumes a continuity that may not be there. We can't evaluate f(2) as the limit t->2- because we haven't demonstrated that f is continuous at t=2.

Solving this problem doesn't have anything to do with limits, or in what order the limits are taken (as was mentioned earlier), the point is that we shouldn't be dealing with limits in the first place. The above paragraph demonstrates the box is empty, and we're done.

knock knock
12-09-2002, 07:05 PM
Cabbage, do you agree that there are infinitely many chips in the first example - when god takes out the highest number in the dish, so that it seems that all the odd integers end up in the dish?

The number of chips in the dish at any time can be given by one piece of information - either it's some nonnegative integer or it's "infinity." Putting 2 chips into the dish means adding 2 to this number, and then taking 1 chip out means subtracting 1 from this number. How can it matter which chip comes out of the dish? You can imagine that god puts two chips in with his right hand, mixes up the chips in the dish, and then pulls one out with his left hand. In one case, this chip happens to be the one with the largest number on it (of the ones remaining in the dish), and in the other, it happens to have the smallest number on it. So what? In one case, this chip had been in the dish for a little while, in the other case, it just went into the dish. So what? We're just counting chips in & chips out, since that's all the affects the current number of chips in the dish. So there should be infinitely many chips in the dish after 2 hours when god takes out the smallest one, just as when he takes out the largest one.

Does anyone here know enough about supernatural numbers to say if that could be a reasonable answer to the question of what chips end up in the dish?

ultrafilter
12-09-2002, 07:22 PM
Once again, if you believe that there are any chips left in the box, name one.

You're not just counting chips here. Your argument that "adding two and subtracting one" leads us to infinity is flawed, because that series is divergent. That doesn't mean that the sum is infinite; that means it doesn't have a sum.

These are not sequences of integers. They're sequences of sets, and as such should not be expected to behave as sequences of integers.

And we probably should be dealing with ordinal numbers to solve this, but that's too complicated, and Cabbage has come up with a simpler proof.

Supernatural numbers don't enter into play here, for two reasons: We haven't all agreed that they exist. Supernatural numbers are a model-theoretic construct, and it's quite (relatively) consistent to assume that there are no such beasts. You'll never break out of the natural numbers by adding one. And that's all you're ever doing--adding one.

Yeah, it's only a two-item list. Tough stuff.

ultrafilter
12-09-2002, 07:25 PM
Originally posted by Chronos
An illustration is in order, here. Consider Graham's Number, a very large number which has come up in theoretical mathematics. We'll need to introduce some new notation here, called arrow notation (I can't make arrows here, so I'll use carats instead). For starters, let 3^3 = 33. Then, let 3^^3 mean 3(33), 3^^^3 = 3(3(33)), etc. OK, now, take 3^3. Take that number, and put that many arrows in the next step, 3^^^^^^^^^^^^^^^^^^^^^^^^^^^3. Now take the number that results from that, and put that in the next step. Continue this for a total of 64 steps, and you've got Graham's Number.

Actually, isn't that just the best-known upper bound for Graham's number?

IIRC, the best-known lower bound on Graham's number is 6.

They're both non-trivial, but not tight.

Tyrrell McAllister
12-09-2002, 08:33 PM
Originally posted by knock knock
We're just counting chips in & chips out, since that's all the affects the current number of chips in the dish. So there should be infinitely many chips in the dish after 2 hours when god takes out the smallest one, just as when he takes out the largest one.

But the very same reasoning argues that the number of chips in the first case (with largest chip removed each time) is zero, because the very simple argument given by several posters in this thread shows that the dish is empty in the second case.

Therefore, since your reasoning gives two different contradictory conclusions depending on what facts you apply it to, you must conclude that this reasoning is fallacious, and abandon it, however attractive it may seem. In particular, you have to abandon the notion that "chips in & chips out" determines the total amount remaining in the end when both "chips in" and "chips out" are infinite. That this reasoning fails is just another peciliarity of infinite sets, such as the fact that an infinite set can be put in one-to-one corresponce with a proper subset of itself.

This is why one has to be careful when applying to infinite cases reasoning that works well in finite cases. It is an at-first-unintuitive fact that some infinite sums can take different values when the terms are rearranged. For example:

1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + . . .
= 1 + 0 + 0 + 0 + . . .
= 1,

even though, just shifting the parentheses over, we can get

(1 - 1) + (1 - 1) + (1 - 1) + . . .
= 0 + 0 + 0 + . . .
= 0.

Cabbage
12-09-2002, 08:56 PM
knock knock, analyzing at each step how many chips are put in, versus how many are taken out, in order to deterine how many chips are left in the box, actually works, but only at the finite level; things get counter intuitive once we bump it up to infinity.

For example, I start with an empty box, I put m chips in the box, then take out n chips (assuming n <= m, both nonnegative integers). How many chips are left in the box? Obviously m-n chips; there's no denying that.

However, in the examples we're looking at, over time, infinitely many chips were put in, and infinitely many were taken out. With this information how can you determine how many chips are left in the box when the whole process is over? The answer is simple--you can't. You need specific information about which chips were put in and which chips were taken out.

Now, thinking of the chips as positive integers again, what if I put all the chips in, and only take out the evens? Then infinitely many chips are left in the box.

What if I put all the chips in, and take all of them out? Clearly none are left in the box.

Now put all of them in, and take out the ones greater than 10. Obviously only 10 are left in the box.

In each case, infinitely many chips were put in, infinitely many were taken out, but the number left in the box is different each time. That's exactly the problem with what's happening in our two examples--in both cases, infinitely many are put in, infinitely many are taken out, but in the first case, infinitely many are left in the box, while in the second case, none are left.

moes lotion
12-10-2002, 12:03 AM
At the risk of flogging the greasy spot on the pavement where the dead horse used to be, let me weigh in on this one more time. I still say that in the second case, the number of chips in the bowl grows without limit. It seems to me that the whole catch to this paradox is the underlying assumption that the process can be completed in a finite time - 2 hours, 2 seconds, whatever. The form of Zeno's paradox to which owlofcreamcheese seems to refer to is the one in which he argues that when a ball is thrown at a wall (say 2 meters away) it can never reach the wall because first it must travel 1/2 way there, the 1/2 the remaining distance, then 1/2 of the remaining remaining distance, and so on. However, legend has it that when Zeno stood in front of the wall and dared all comers to try to hit him he was forced to call off the demonstration due to the nosebleed he got when the first nay-sayer nailed him. Clearly, even though the ball must cross an infinite number of ever shorter distances, the total distance is finite.

And Cabbage, I'll name one just as soon as you name the largest integer!

Cabbage
12-10-2002, 01:14 AM
knock knock, how would you analyze the following two scenarios:

(In both cases, we're using the same division of the two hours into steps as we've been doing throughout the thread).

1. First step, God throws the number 1 in the box. Second step, God throws the number 2 in the box, and then removes the largest number from the box (2 at this stage). At each of the remaining steps, God always throws the next number in the box (3,4,5,...), then removes the largest number from the box.

2. First step, God throws the number 1 in the box. Second step, God throws the number 2 in the box, and then removes the smallest number from the box (1 at this stage). At each of the remaining steps, God always throws the next number in the box (3,4,5,...), then removes the smallest number from the box.

How many numbers are left in the box after the completion of the steps in case 1? Case 2? Be sure to note that both cases are exactly the same in the sense you've mentioned above--in both cases, God simply throws a chip in on the first step. And, again in both cases, for each of the following steps, he adds a chip, then removes a chip. By your reasoning, therefore, both boxes should wind up with the same number of chips when all is said and done. Is this so in these two cases?

moes lotion, what you've posted seems to have everything to do with the way we've divided up the two hours, but nothing to do with how many chips are left in the box.
And Cabbage, I'll name one just as soon as you name the largest integer!
Ah...proof by intimidation. ;)

12-10-2002, 02:14 AM
OK,

Step 1: God puts chip #1 in the bowl.

Step 2: God puts chip #2 in the bowl.

Step 3: God takes chip #1 out of the bowl.

Elapsed time, one hour, chip #2 is in the bowl.

Step 4: Chip #3 in.

Step 5: Chip #4 in.

Step 6: Chip #2 out.

Elapsed time, one hour and thirty minutes, Chips #3 and #4 are in the bowl.

Is there a rule that says that at two hours, after an infinite number of iterations that God must have completed some number of steps that is exactly divisible by three? Why?

OK, we ask God to do each set of three steps as a single operation in a finite, but always decreasing time interval, even dividing intervals less than the Planck interval, and approaching two hours as a limit.

In His infinite wisdom, and mercy, God does as we ask. (I certainly wouldn't.)

So, there are always 2n - n chips in the bowl. At the end of the time period whose limit is defined by the function for speed, that number n will be infinite, but will be a member of the set of natural numbers. The answer is there are n chips in the bowl, and the value of n is infinite. The least value in the bowl, is the chip numbered n. There is no finite answer for the value of n, and no finite number on any of the chips in the bowl. All of the chips in the bowl will have values inscribed on them, which are infinite integers. We cannot name them, but that is not a problem for God. We were the ones who asked Him to do this silly experiment.

So, asking for the name of an infinite integer is no less silly if you ask God to do it, than it is if you ask some mathematician.

Tris

12-10-2002, 02:19 AM
Oops, the smallest value would be n + 1, not n.

Sheesh. Off by one errors, at infinite magnitutes.

Tris

Cabbage
12-10-2002, 03:15 AM
And Cabbage, I'll name one just as soon as you name the largest integer!
Actually, I guess I should respond to this, rather than just dismiss it, as I did above; I don't know whether the quote was meant facetiously or not.

The difference here is that I never claimed there is a largest integer; in fact, I'll go on record here and state that there is no largest integer. Why ask me to name something when I deny its existence?

On the other hand, you do, in fact, claim there is a number in the box (infinitely many of them, in fact). It's perfectly reasonable for me to ask you to name one of them if you claim that such a thing exists. If, for whatever reason, you claim to not be able to name one, yet still maintain that such a thing exists, then give me an argument on how this can be so. Why can't you name one?

Triskadecamus, sure, at each step, there are 2n - n chips in the box. But that only works so long as only finitely many steps have been completed. Give me some justification that you can use similar logic to deduce how many chips are left after infinitely many steps have been concluded.

Consider this--say we start with a box already full of all the positive integers (and only the positive intgers). At each step, I remove the smallest one, leaving infinitely many. How many are left at the end of the two hours now? Again, it should be clear that none are left (if any are left, consider the smallest one left--ah, but it would have been removed in the step immediately following when its predecessor was removed--Contradicting the assumption that it was left), even though at every individual step, infinitely many are in the box.

Now if you buy that argument, do you see that it must necessarily imply that no chips are left in the original example (where God removes the smallest chip at each step, after throwing two chips in). In the original example, God is putting all of the positive integers in (and only the positive integers) two at a time. Now, in my newest example, I'm saying the hell with it, just throw them all in there to begin with (but do everything else the same). Now remove them one at a time--removing the smallest one at each step. There won't be any left--I've removed them all, dammit! And neither can there be any left in the original example.

Achernar
12-10-2002, 05:00 AM
Nothing good to add, but when I first read about this paradox, it was explained by the sentence, "We have no mathematics to describe the completion of infinite tasks." I suppose it's rather like saying, "Count 1, 2, 3,... What number do you finish on?" The fact that it takes a finite time changes nothing.

Scuba_Ben
12-10-2002, 08:18 AM
Originally posted by KidCharlemagne
Aren't different size infinities called alephs or something?

Cardinalities of infinities are represented by the Hebrew letter aleph. (I'd include the letter, but I don't know how to hamsterize the VB code to do so.) The set of integers, the "countable" infinity, is aleph0; the set of reals is aleph1; the set of (I think) curves is aleph2.

I learned this from One, Two, Three... Infinity, which only went that far on infinities. Conceptually, there would be an infinite number of infinities, but my mind seg faults on the idea of aleph2. Professional mathematicians are welcome to discuss higher order infinities.

bryanmcc
12-10-2002, 08:38 AM
Perhaps this will help: God has a bowl with zero chips in it, and puts in (-2) more. How many are in the bowl now? Well, -2, of course. But how do you put -2 chips in? What the hell are negative chips? Of course there is no such thing, and doing so is imposible. This is why we assigned the action to God in the first place: he is a being by definition capable of doing the impossible. So he is going to be the one to put infinitely many chips in and take infinitely many out in two hours, because doing so is obviously, blatantly impossible. We are not surprised that regular addition of positive integers doesn't help when God puts in -2 chips; likewise we should not be surprised when finite arithmetic doesn't tell us what happens when infinitely many chips are added and removed. And we also shouldn't be surprised when our common sense gives the wrong answer in such unfamiliar cases. _If_ we accept that God is doing the impossible, then we must accept that there are -2 chips in the bowl in my example, or zero chips in the bowl in the OP's example.

ZenBeam
12-10-2002, 11:28 AM
Suppose each chip has a different thickness. In particular, suppose the thickness of chip k is 1/k. God, with His infinite sense of aesthetics, doesn't just toss the chips into the dish. He stacks them up. The height of the stack after step N is Sum [k = 1 to N] {1/(2*k-1) + 1/(2*k )- 1/k }. After 2 hours, the height of the stack would be Sum [k = 1 to infinity] {1/(2*k-1) + 1/(2*k) - 1/k } = Sum [k = 1 to infinity] {1/(2*k*(2*k-1)) } This sum converges to a finite number greater than zero, call it S. Are we now saying this sum does not equal S?

I have a difficulty reconciling a stack of height greater than zero, made up of zero chips. I also have a difficulty believing that Sum [k = 1 to infinity] {1/(2*k*(2*k-1)) } approaches S, but is in fact equal to zero. I would not have a problem if the number of chips were taken to be undefined or indeterminate, but no one seems to be arguing that.

Note that I get reasonable answers to the height of the stack for both of these scenarios from cabbage:
1. First step, God throws the number 1 in the box. Second step, God throws the number 2 in the box, and then removes the largest number from the box (2 at this stage). At each of the remaining steps, God always throws the next number in the box (3,4,5,...), then removes the largest number from the box.

2. First step, God throws the number 1 in the box. Second step, God throws the number 2 in the box, and then removes the smallest number from the box (1 at this stage). At each of the remaining steps, God always throws the next number in the box (3,4,5,...), then removes the smallest number from the box.The height of the stack for the first case approaches 1, the height of the stack for the second case approaches zero.

Cabbage
12-10-2002, 02:01 PM
Scuba_Ben
The set of integers, the "countable" infinity, is aleph0; the set of reals is aleph1; the set of (I think) curves is aleph2.
That's the common "continuum hypothesis" mistake. True, the set of integers is aleph0. The set of reals is bigger than aleph0, but it may also be much bigger than aleph1, as well. I'm not sure what you mean by curves, but I figure either the set of functions from the reals to the reals, in which case that would certainly be a set bigger than the reals, and not necessarily aleph2. I'm not sure off hand if the set of continuous functions from the reals to the reals is strictly larger than the reals, but it can still certainly be much bigger than aleph2, as well (since the reals can be much bigger than aleph2).

ZenBeam, you're making the same mistake I accused knock knock of making when I posted after your earlier post--you're assuming a continuity that isn't there. Your sum is meaningful so long as there are finitely many chips in the box, but you haven't demonstrated that it still has any meaning when we take it to the infinite level.

Consider this classic example: I don't want to get bogged down in the notation and algebra, so I'll simply describe the following sequence of functions.

The nth function fn (n=1,2,3,...) mapping the interval [0,1) is described as follows. It's zero from x=0 until we get to the point x= 1 - 1/n, at which point it takes off in a straight line up to the point (1, 2n2).

So, in general, the function fn looks flat for a considerable length, then suddenly shoots up steeply. What is the integral (area under the curve) of fn? Well, it gets no area for the flat part; when it shoots up steeply it forms a triangle with base 1/n and height 2n2. This triangle has area n.

These functions converge pointwise to another function, call it finf. What is the integral of this function? The integrals of the functions in the sequence are growing without bound, so by your logic, the integral of finf should be infinite.

However, for any x you pick in [0,1), the sequence fn(x) converges to zero. So actually, finf is identically zero on [0,1}, hence the integral of finf is zero.

Similarly, despite the fact that the chips in the box are growing without bound throughout the two hours, you just can't rely on limits and convergence to tell you how many chips are in the box at the end of the two hours. What is your assumption that there is any convergence to begin with based on?

moes lotion
12-10-2002, 03:40 PM
Okay, here is my reasoning for saying that the number of chips in the bowl grows without limit - which is equivalent to saying that the bowl is never empty. To re-iterate the process:

step 1 - place chips 1 & 2 in the bowl
step 2 - place chips 3 & 4 in the bowl and remove chip 1
step 3 - place chips 5 & 6 in the bowl and remove chip 2
.
.
step n - place chips 2n - 1 and 2n in the bowl and remove chip n - 1

Note that for each step, the number of chips in the bowl is exactly n + 1.

Now, let n approach infinity. QED

ultrafilter
12-10-2002, 03:43 PM
Originally posted by moes lotion
Okay, here is my reasoning for saying that the number of chips in the bowl grows without limit - which is equivalent to saying that the bowl is never empty.

You're right. After any finite number of steps, the bowl is never empty. That doesn't mean that it isn't empty after an infinite number of steps.

ultrafilter
12-10-2002, 03:46 PM
Originally posted by Cabbage
I'm not sure off hand if the set of continuous functions from the reals to the reals is strictly larger than the reals...

They're the same size. A continuous function from R to R is essentially a function from Q to Q. It's not hard to show that Q -> Q is the same size as R, but I'd have to work out the details again.

ultrafilter
12-10-2002, 03:48 PM
btw, y'all might be interested in this thread (http://boards.straightdope.com/sdmb/showthread.php?s=&postid=2696584) I started on large numbers. Chronos's discussion of Graham's number got me thinking, and I ran from there....

William_Ashbless
12-10-2002, 03:48 PM
Originally posted by moes lotion
Okay, here is my reasoning for saying that the number of chips in the bowl grows without limit - which is equivalent to saying that the bowl is never empty. To re-iterate the process:

step 1 - place chips 1 & 2 in the bowl
step 2 - place chips 3 & 4 in the bowl and remove chip 1
step 3 - place chips 5 & 6 in the bowl and remove chip 2
.
.
step n - place chips 2n - 1 and 2n in the bowl and remove chip n - 1

Note that for each step, the number of chips in the bowl is exactly n + 1.

Now, let n approach infinity. QED

Except that, as has already been mentioned, any particular finitely numbered chip has already been removed. Pick one. 147? 20045? 999998? They've all been removed, and we can tell you exactly at what step. That is, you cannot specify a particular chip that has not been removed.

If all of the finitely-numbered chips have been removed, you're not really left with much.

moes lotion
12-10-2002, 04:00 PM
I promise that this is my last response to this thread. ultrafilter, lets play the following game of leap frog. You name a chip that's been removed from the bowl, call it n. Now, on my turn I say fine, when chip n was removed on turn n + 1, chips 2n - 2 and 2n - 1 were put in the bowl, when were they removed? Your turn...

Note: the first one to stop loses.[smileyface]

ultrafilter
12-10-2002, 04:20 PM
Originally posted by moes lotion
I promise that this is my last response to this thread. ultrafilter, lets play the following game of leap frog. You name a chip that's been removed from the bowl, call it n. Now, on my turn I say fine, when chip n was removed on turn n + 1, chips 2n - 2 and 2n - 1 were put in the bowl, when were they removed? Your turn...

Note: the first one to stop loses.[smileyface]

You're assuming that this function is continuous at infinity (which is a meaningless phrase to begin with). We can go on forever, but the behavior of this process after any finite number of steps implies nothing about its behavior after an infinite number of steps.

Tyrrell McAllister
12-10-2002, 04:33 PM
Moving on, it is, to me at least, an interesting paradox to note that the following problem is ill-posed.
Suppose that at time t = 0 a dish is empty, and that for each non-negative integer N, at time t = 2 - (1/2)N, God takes two chips from a cache of identical chips, and simultaneously places them into the dish and removes one of the chips in the dish. How many chips are in the dish at time t = 2?Somehow, in what order the chips are removed matters, even though they are all identical. The universe, as it were, knows in what order the chips are being placed into the dish, and, dispite their identicalness, keeps track of when each is removed. For, after all, something has to happen at the two-hour mark, and the only way for the universe to know what to do is to keep track of that information.

In other words, the problem is ill-posed because it doesn't mention the additional presence of a sort of "tagging" operation that would be occuring, were the experiment actually performed. Even though the chips are, in all respects, indistinguishable, the universe is saying "for the duration of this experiment, you're chip 1, you're chip 2, ..." and so on. But there are degrees of freedom to this hidden tagging operation that make different outcomes possible, depending on how the tagging operation is performed in a particular case.

Perhaps this is a clue to a solution to the paradox with the lightbulb being turned on and off at Zeno-type intervals. Is the light on or off at the 2 hour mark? This phrasing of the question seems not to determine an answer, but perhaps, in each particular case where this procedure is performed, a similar unmentioned "tagging" operation is performed in one of several possible ways, and the particular way in which it is performed does suffice to determine the outcome.

Cabbage
12-10-2002, 04:38 PM
Here's another classic example to think about:

Imagine a square, each side one unit long. Consider two opposite corners of the square, say the bottom left corner and the top right corner. Imagine there's a "stairway" from this bottom corner to the top corner. Say it has n steps, and the steps are evenly spaced, as staircases usually are. What is the length of this staircase? Well, all the horizontal segments add up to one unit, as do the vertical segments, for a total of two units.

What happens as we let the number of steps increase, without bound? Well, the length of the staircase as still two units. Additionaly, the staircase begins to approximate the diagonal connecting the two corners, with greater and greater precision as the number of steps increase.

By the logic being used to show there are infinitely many chips in the box, we can conclude that the length of the diagonal is equal to the limit of the length of the staircase, as the number of steps increases.

And we've just proven that 2 = sqrt(2).

The lesson to be learned--limits don't preserve everything. In particular, limits don't preserve the number of chips in the bowl in the example we've been considering.

William_Ashbless
12-10-2002, 04:46 PM
Naturally, at any stage in the staircase process, it is true that the distance between two points that are along the diagonal line between two steps is sqrt(dx*dx+dy*dy) where dx is the x delta, and dy is the y delta. Since dx = dy, this is sqrt(2dx) of course, or sqrt(2)sqrt(dx), and if you sum all those up over the width of the cube, you get sqrt(2). This must remain true even if you have stairs that approach being smooth.

While the staircase approximates the diagonal, at any particular finite stage dx+dy (or 2dx since dx = dy) never really represents the distance along the diagonal even though the staircase approaches a smooth diagonal.

This is an excellent demonstration of the fallibility of limits -- one has to be terribly careful.

William_Ashbless
12-10-2002, 04:53 PM
oops, I meant since dx=dy, this is sqrt(2dx*dx) which ends up being dx * sqrt(2), but the rest of it applies.

ZenBeam
12-10-2002, 05:09 PM
What is your assumption that there is any convergence to begin with based on?At each step, the thickness of the stack increases by 1/(2*k*(2*k-1). For any epsilon > 0, I can give you a step N beyond which the sum is always within epsilon of S (= log(2)). Isn't this (more or less) the definition of convergence?

Your sum is meaningful so long as there are finitely many chips in the box, but you haven't demonstrated that it still has any meaning when we take it to the infinite level.Does not this argument apply to any infinite sum? How can I ever say any infinite sum, e.g. Sum [k = 1 to infinity] {(-1)k+1 / k }, has a value? My book says this is equal to log(2), and I bet you've got more than one book which agrees with this. How would you "demonstrate that it still has any meaning when we take it to the infinite level"?

If the chips are not numbered, does the height of the stack converge to a value greater than zero? If instead of summing chip thicknesses, I sum line lengths in the same fashion, certainly you agree that converges, don't you?

Cabbage
12-10-2002, 05:49 PM
ZenBeamIsn't this (more or less) the definition of convergence?
Yes, but what's missing is whether or not this is an appropriate way to describe the height of the chips after the full two hours is up.
...My book says this is equal to log(2), and I bet you've got more than one book which agrees with this. How would you "demonstrate that it still has any meaning when we take it to the infinite level"?
It's important to remember that we're not actually adding up infinitely many numbers. The sum of the infinite series means that, as we add up the terms of the series (in that particular order), the sum gets arbitrarily close to log(2). That's important in some cases. For example, if you wanted a decimal expression for log(2), you could sum up some initial terms in that series--the farther you go out, the closer you'll be--that would be a nice application of an infinite series.

I should mention here yet another counter-intuitive aspect of infinity. About that infinite series just mentioned above--pick a real number. I can rearrange the terms of that series so that it will converge to the real number you picked. Again, order is important here.

In the original example, at each step the box contains certain integers; after all the steps have been taken, the box will still contain a subset of the integers. The limit of the number of chips in the intermediate steps is not the same as the number of chips in the box in its final state--it's that simple.
If the chips are not numbered, does the height of the stack converge to a value greater than zero? If instead of summing chip thicknesses, I sum line lengths in the same fashion, certainly you agree that converges, don't you?
If you're talking about adding chips and removing chips along the way, then no there's no way to talk about convergence in any way. Remember, the ordering of the chips is important, just as the ordering of the above infinite series is important. You might add and remove the exact same chips, only in a different order than before, and the height of the stack will converge to a completely different value.

Same thing with line lengths, if we go along and say something to the effect, "Then at this step I add in a line segment of length one, and take away a line segment of length 1/2, then add a line segment of length 2, take away a length of 2.3,...", then it's absolutely impossible to say anything about the what the length will be in the end. And if you do the same steps in a different order, you might get it converging to a different value (just as the infinite series demonstrates).
They're the same size. A continuous function from R to R is essentially a function from Q to Q. It's not hard to show that Q -> Q is the same size as R, but I'd have to work out the details again.
That's right, I see what you're saying. Thanks, ultrafilter.

ultrafilter
12-10-2002, 05:51 PM
Originally posted by Cabbage
That's right, I see what you're saying. Thanks, ultrafilter.

No problem.

Tyrrell McAllister
12-10-2002, 05:53 PM
Even if a function f(x) converges to a particular value y0 as x approaches a point x0, it needn't follow that f(x0) = y0.

So, in the case of you height function H, given by
H(N) = Sum [k = 1 to N] {1/(2*k-1) + 1/(2*k )- 1/k },we do indeed have that H(N) converges to some nonzero number y0 as N goes to infinity, and by a certain abuse of notation, we could write
H(infinity) = limit [N --> infinity] H(N) = y0.Now, by doing this, we would be defining H(infinity) to be the limit of a certain sequence of partial sums. But there is no guarantee that this definition would match the more natural definition of H(infinity), which is as the height of the stack of chips in the dish at t = 2 hours. And, in fact, under this second definition, H(infinity) = 0.

moes lotion
12-11-2002, 12:07 AM
Okay, I was wrong - about my last post being the last post I'll make in this thread. Once more into the breech...

It seems to me that the underlying sticking point here is that I don't accept that the process as described can ever stop. Because the set of +ve integers is infinite (a point on which I trust all agree), therefore there is no last step - the whole argument about what's in the bowl after the deity of your choice has finished simply doesn't apply.

ultrafilter said
You're right. After any finite number of steps, the bowl is never empty. That doesn't mean that it isn't empty after an infinite number of steps.

But the limit as n aproaches infinity of 1 + n is infinity. Surely you don't have trouble understanding this? I'm not saying that is is a particular number, but that is is larger than any number you can name, including 0.

William_Ashbless said
Except that, as has already been mentioned, any particular finitely numbered chip has already been removed. Pick one. 147? 20045? 999998? They've all been removed, and we can tell you exactly at what step.

Alright, at which step was the last chip in the bowl removed? You can't answer this because there is no such chip. The number of chips in the bowl at each step increases. There is no last step. The bowl is never empty. So there.

And that is the last I have to say here.

t-keela
12-11-2002, 12:23 AM
I've been away for awhile, (at least from this thread). So, what'd ya'll come up with?

Did y'all reach some kind of agreement or what? I don't know if I even remeber the OP at this point, but IIRC dish #1 was the infinite set of chips and ..dish #2 was the I set of chips minus 1, right?

That makes one set all even # chips and one set all odd # chips. Which is in fact a set of two which can be described as (n, n +1....to infinity).

Did I miss something or is someone attempting to say there are no chips in the first dish?

While I might be willing to ponder the idea of there only being one chip in set one...I don't think I'll agree with zero chips being in set one.

anyway, just thought I'd check in and see how it was going with you diehards...:D

Tyrrell McAllister
12-11-2002, 01:54 AM
Originally posted by moes lotion
Alright, at which step was the last chip in the bowl removed? You can't answer this because there is no such chip. The number of chips in the bowl at each step increases. There is no last step. The bowl is never empty.
Everything you say here is true, except for the very last sentence. That conclusion simply does not follow. If such reasoning were valid, Zeno's paradoxes would prove the impossibility of motion. If I leave my room, there may be no last point that I pass through before I reach the door, yet still I do pass through allthe points between here and the door, and in finite time to boot.

I'll give a situation more similar to the one under discussion, to show that there being no last chip proves nothing. Consider this thought experiment. Two dishes lie before God. At the beginning, dish A contains infinitely many chips, each labeled with a unique one of the nonnegative integers, and dish B is empty. For each nonnegative integer N, at time t = 2 - (1/2)N hours, God takes the chip labeled N from dish A and puts it in dish B.

With this setup, you'll have to agree, every chip is taken from dish A and placed in dish B before 2 hours have passed. Therefore, after 2 hours have passed, dish A will be empty. This dispite the fact that there was no last chip removed from dish A.

12-11-2002, 02:03 AM
If God threw the removed disks into another bowl, would it be empty too? Would it have more disks that the first bowl, or less? Why? Doesn't it get disks at the exact same rate? So, an infinite series of steps which is stated to have ended must have an infinite number of steps, but it must also be ended, and therefore must have a maximal value, a unique number. But the definition of infinite denies that it can have such a value. Now you want to know the value.

This insistence upon naming a specific transfinite integer is silliness. OK, let's ask God. "What is the lowest number?" God says, "Gazornipleximan." "But wouldn't that one have been removed after you had done gazornipleximan steps?" We ask. "Gazornipleximan is an infinite number, two times Gazornipleximan is still Gazornipleximan. Not only that, but Gazornipleximan over two, minus one is also equal to Gazornipleximan." But you can't have numbers that work like that! We howl. "Yes, I can, I'm God, remember? That's why you needed me to do this silly trick in the first place, because no one else can do something infinite and be done at a specific time. When you do that, all the numbers come out Gazornipleximan."

Infinite doesn't have a specific value. Asking for the specific integer value of a number stated to be developed by an infinite series is just a silly question. It's infinite. No, the arithmetic doesn't work, because it's infinite. This is sort of why infinite shit doesn't happen with integers in physical phenomena.

Tris

RM Mentock
12-11-2002, 02:56 AM
Originally posted by moes lotion
The number of chips in the bowl at each step increases. There is no last step. The bowl is never empty. So there.
So there?

OK I give up.

William_Ashbless
12-11-2002, 08:56 AM
If God threw the removed disks into another bowl, would it be empty too? Would it have more disks that the first bowl, or less? Why? Doesn't it get disks at the exact same rate? So, an infinite series of steps which is stated to have ended must have an infinite number of steps, but it must also be ended, and therefore must have a maximal value, a unique number. But the definition of infinite denies that it can have such a value. Now you want to know the value.

This insistence upon naming a specific transfinite integer is silliness. OK, let's ask God. "What is the lowest number?" God says, "Gazornipleximan." "But wouldn't that one have been removed after you had done gazornipleximan steps?" We ask. "Gazornipleximan is an infinite number, two times Gazornipleximan is still Gazornipleximan. Not only that, but Gazornipleximan over two, minus one is also equal to Gazornipleximan." But you can't have numbers that work like that! We howl. "Yes, I can, I'm God, remember? That's why you needed me to do this silly trick in the first place, because no one else can do something infinite and be done at a specific time. When you do that, all the numbers come out Gazornipleximan."

Infinite doesn't have a specific value. Asking for the specific integer value of a number stated to be developed by an infinite series is just a silly question. It's infinite. No, the arithmetic doesn't work, because it's infinite. This is sort of why infinite shit doesn't happen with integers in physical phenomena.

Tris

The very fact that 'infinite doesn't have a specific value' immediately suggests that there isn't a specific chip in the bowl. If, at any point of time, including, if you wish, the point at the two hour mark, you look at the bowl, you can only say 'there are this many chips' is if you can actually look and see that a chip is there. Any chip.

You can easily do this at any point up to but not including the two hour mark, knowing full well that any chip you can describe that you see will be gone in a finite period of time.

At the two hour mark, there is no way to enumerate, no way to describe, no way to see, no way to point to, no way to discuss any coin or coins in the dish. Whether you believe that naming a specific one is silliness or not, every actual single coin ever involved in the problem is given a unique integer by the definition of the problem, and coin i has already been removed.

If you start talking about the coin of index Gazornipleximan, and Gazornipleximan is 'an infinite number', and infinite numbers are not specifically an integer, then which coin are you referring to? If you're not referring to a specific coin, then the fact that you've given it a name is irrelevant -- it could refer to any coin. Using an index that cannot possibly refer to a specific coin to suggest that that coin is in the dish is, to me, the true silliness. There's no justification.

erislover
12-11-2002, 08:58 AM
Appealing to the ineffability of infinity is not very convincing. It seems perfectly reasonable that one bowl is empty and the other isn't.

Give god an infinite number of bowls. Have him take out the Nth prime at time N and all of its multiples, and put them in one of his infinite bowls (which are marked 2, 3, 5, 7, 11, and so on). Now he has an infinite number of bowls with an infinte number of chips. (given the time formula as before etc). This is damned counter-intuitive, but this is how we do math.

That is simply what happens. Because we are dealing with infinity, the rate at which we approach it is irrelevent. It takes just as long to get there if we count by 1s, 2s, or primes, and we pass just as many numbers. Only, if we are removing some on the way, the manner in which we do so will determine if any, and how many, are left.

Can I get a hell yeah?

William_Ashbless
12-11-2002, 09:05 AM
Works for me. Infinity is counterintuitive, and despite objections otherwise, we've established some of these counterintuitive 'truths' and 'rules' to help us deal with goofy things like Zeno's Paradox, converging geometric sums, and integration near one of these infinite points.

Integrating under the curve f(x)=1/(x^2) is always a good example... Chaos to those who try to evaluate f(x) at point 0, but we can still compute the area under the curve in a range of x that lies across this point.

knock knock
12-11-2002, 11:22 AM
What if god decided to pursue a slightly different strategy? Instead of putting the 2 smallest remaining chips into the dish, he put the 2 smallest remaining even chips in, and if there were no even chips left, then he'd put the 2 smallest remaining odd chips in. Then he'd remove the smallest chip (perhaps with the same preference for even chips?) from the dish as before. So, he'd put in 2,4, take out 2, put in 6,8, take out 4, etc. Perhaps, at some point, all the even chips would be in the dish or already taken out, so he would put in the 1,3, etc.

If he followed the same time pattern, where would the chips be after 2 hours? Case 1: if he takes out the smallest even chip, and only removes an odd chip if there are no even chips to remove, then, assuming ultrafilter's arguments to be correct, all the even chips would have been taken out of the bowl. But where would the odd chips be? Are they all in the bowl? But when did he put them in? Did they all go in instantaneously at the very end of the time limit? Or not yet in the bowl? But then why did he stop putting chips into the bowl if there were still some left? Or all taken out of the bowl? But then when did he remove them? Wasn't he always in the process of removing even chips? Did they all go in and out instantaneously?

Case 2: he removed the smallest chip from the bowl, whether it was even or odd. Similar questions apply for the odd chips, and the state of the even chips is also not clear.

Is there an answer in these situations? Are the answers different? Does the question even make sense?

ultrafilter
12-11-2002, 11:26 AM
In that last case, no odd chip will ever be put in the bowl, because there will always be more even chips left to put in.

All of these problems are good examples of why we don't talk about completing infinite processes. There are just too many contradictions.

William_Ashbless
12-11-2002, 12:08 PM
I agree with ultrafilter.... 'Enter At Your Own Risk', and do so treading lightly.

knock knock points out an interesting problem. Strictly speaking, there is no time period before two hours when all the even coins are gone. After the two hour mark it cannot reasonably be said that even coins remain, since no distinct coin can have been left behind. However, since this condition only 'exists' instantly following the two hour mark, there has been no time to begin putting in or taking out the odd chips. Since the two hour mark ends the experiment, the even chips have been put in and taken out, but the odd chips remain untouched, never entering the dish.

That's my take on it anyway.

William_Ashbless
12-11-2002, 12:10 PM
sorry, I meant 'since no distinct even coin can have been left behind', but I'm sure you parsed that.

erislover
12-11-2002, 12:46 PM
The result doesn't sound so silly if the criteria for removing chips in the OP is changed.

"Place an infinite amount of chips in. Remove odd chips. How many remain?"

vs

"Place an infinite amount of chips in. Remove all chips. How many remain?"

The trick is like the "where's the missing dollar"... that is, in the presentation.

ZenBeam
12-11-2002, 01:10 PM
First, three equations. I'll assume these summations are carried out in 2 hours in the fashion of the OP, although that really isn't relevent for most of this post.

Eq. (1):
S1 = a + b - a + c + d - b + e + f - c + ...

Eq. (2):
S2 = a - a + b - b + c - c + d - d + ...

where Eq. (1) is the summation we're interested in and Eq. (2) is obtained be rearranging the elements of Eq. (1). S2 = S1 if the summation in Eq. (1) is absolutely convergent (http://mathworld.wolfram.com/AbsoluteConvergence.html). If the summation in Eq. (1) is not absolutely convergent, S2 is not in general equal to S1. Eq. (1) is absolutely convergent if the sum of the absolute values of all the terms converges, i.e. the following summation is convergent:

Eq. (3):
S3 = |a| + |b| + |-a| + |c| + |d| + |-b| + |e| + |f| + |-c| + ...

Now, four examples:
Example 1:
a = 1, b = 1/2, c = 1/4, d = 1/8, and so forth. a, b, c, etc. could represent chip thicknesses in the OP, or the distance an infinitesimally small fly travels, beginning at X = 0. In this example, Eq. (1) is absolutely convergent, and Eq. (2) can be used. You can also sum the terms in Eq. (1), and you get the same answer. The fly ends up at X = 0, where it started.

Example 2:
a = 1, b = 1/2, c = 1/3, d = 1/4, and so forth. a, b, c, etc. can again represent chip thicknesses in the OP (as in one of my previous posts), or the distance an infinitesimally small fly travels, beginning at X = 0. In this example, Eq. (1) is Conditionally Convergent (and note Eq. (1) on this page) (http://mathworld.wolfram.com/ConditionalConvergence.html), and Eq. (2) can not be used. You can, however sum the terms in Eq. (1) and the answer converges. The fly ends up at X = log(2).

Example 3:
a = 1, b = 1, c = 1, d = 1, and so forth. a, b, c, etc. can again represent adding chips in the OP (or removing them for the terms with the minus sign), or again the distance an infinitesimally small fly travels, beginning at X = 0. In this example, Eq. (1) is divergent (http://mathworld.wolfram.com/DivergentSeries.html), and again Eq. (2) can not be used. Summing up the terms, the fly heads off in a jerky fashion towards infinity.

Example 4:
a = Chip 1, b = Chip 2, c = Chip 3, d = Chip 4, and so forth. For "Chip 1", "Chip 2", etc., you can use sets, or unit vectors in chip space, or however you want to express it. The partial sums give the chips in the dish after each step in the OP. In this example, Eq. (1) is again divergent, and again Eq. (2) can not be used.

When the argument is made (and I'll paraphrase, since it's been made many times) that there are zero chips in the dish, or that every chip in the dish has been removed, the writer is implicitly using Eq. (2) to solve Example 3 or 4. By using words instead of equations, it isn't obvious, but the pairing of a with -a, b with -b, and so forth, as in Eq. (2), is always made. For examples 2 through 4, this is invalid. You simply can't validly rearrange the terms to pair off each term with its negation if the series is not absolutely convergent.

Everyone should be able to convince themselves of this by applying the same argument to example 2: For every step where the fly flies in the plus direction, there is a corresponding flight in the -x direction. There is no flight in the +x direction not canceled by a similar flight in the -x direction. Hence, the fly ends up at the origin. This argument is clearly false for example 2; the distance between the fly and log(2) is bounded closer and closer to zero as t --> 2 hours. The fly ends up at X = log(2).

I'll also address the "name one" argument in two ways. First, I think everyone agrees that the limit, as t --> 2 hours, of the number of chips in the dish is infinite, even if we disagree on the number at t = 2 hours. Even so, in the limit t --> 2 hours, no chip can be named that is in the dish.

Secondly, consider example 2 again. For the fly to end up at log(2), instead of the origin, some flight in the +X direction must not have been canceled by a flight in the -X direction. No such flight can be named, but there's the fly at X = log(2).

Arguments have been made that the number of chips in the dish is not necessarily continuous at t = 2 hours, but without the above arguments for why the number of chips in the dish must be zero, this argument doesn't lead anywhere.

The "name one" argument, and the arguments pairing each chip with its removal are, I believe, the only arguments in this thread for there being zero chips in the dish at t >= 2 hours (although it is a long thread). Since these argments are invalid, there is no justification for saying there are zero chips in the dish. Now, if someone can show, using valid equations and valid manipulations of those equations that the sums in examples 3 and 4 are zero, I'd like to see that. I haven't yet.

ultrafilter
12-11-2002, 01:36 PM
Equations aren't what matters here. The order in which you remove chips is what matters.

A different example is in order. Suppose I'm playing at a slot machine, with a countably infinite number of quarters. Each time I play, I win a countably infinite number of quarters. As above, let's assume that the kth play takes 1/k minutes.

If I just keep playing the quarters I started with, I'll always have a countably infinite number of quarters. But if I dovetail, I can be out of quarters at the end of 2 minutes.

I play the same number of quarters in each scenario. The only difference is the order in which I play them.

ZenBeam
12-11-2002, 01:43 PM
Equations aren't what matters here. The order in which you remove chips is what matters.Equations are how you ensure you are removing chips in the right order.

ultrafilter
12-11-2002, 01:50 PM
Originally posted by ZenBeam
Equations are how you ensure you are removing chips in the right order.

No. What ensures you're removing chips in the right order is the sequence of sets of chips in the bowl after each step. All the equations can do is track the cardinality, which isn't enough information to determine the behavior of the function at infinity.

I hate to use that term when it has such potential for confusion, but I really can't think of anything else.

Tyrrell McAllister
12-11-2002, 02:28 PM
ZenBeam, please see my post (http://boards.straightdope.com/sdmb/showthread.php?postid=2699024#post2699024) from the previous page, where I point out that we (those of us who argue that the dish is empty at two hours) are not evaluating a series, implicitly or otherwise, to arrive at our conclusion. Nothing so sophisticated as infinite series are necessary. The problem can be analysed with simple set threory (and if set theoretic reasoning is invalid, then the entire ediface of mathematics, including the mathematics of series evaluations, is worthless).

I will try to make the set theoretic reasoning more explicit.

Let N be the set of positive integers. Given any subset S of N, denote by Sc the complement of S in N; that is, Sc is the set of all those numbers in N that are not in S.

Lemma:Given any subset S of N, if, for each number n in N, we have that n is in Sc, then S is empty.
Proof: This is a trivial proof by contradiction. Suppose that S is not empty. Then S contains some positive integer, say m. By hypothesis, every positive integer, including m, is in Sc. Therefore, by the definition of Sc, m is not in S, a contradiction.

In the theorem that follows, I'm going to identify each chip with the number written on it. So now N may be thought of as the set of all the chips.

Theorem:The set D of chips in the dish at two hours is empty.
Proof: We show that each chip is in Dc; the conclusion will then follow from the Lemma. In principle, Dc contains two kinds of chips:
(i)those chips that aren't placed in the dish before the two hour mark, and
(ii)those chips that are removed from the dish at a time before the two hour mark.I think that everyone here agrees that the set of chips satisfying condition (i) is empty. That is, every chip is eventually placed in the dish at some time before the 2 hour mark. So the problem reduces to showing that every chip in N is of type (ii).

To prove this, suppose that n is a chip in N. Then n is removed from the bowl at time tn = 2 - (1/2)n-1. Note that, regardless of the value of n, we have that tn < 2. Therefore, chip n is removed from the dish before the two hour mark, so chip n is in Dc.

Since we showed this for an arbitrary chip n in N, we've shown that every chip in N is in Dc. Therefore, by the Lemma, D, the set of chips in the dish, is empty.

knock knock
12-11-2002, 04:25 PM
Originally posted by William_Ashbless
I agree with ultrafilter.... 'Enter At Your Own Risk', and do so treading lightly.

knock knock points out an interesting problem. Strictly speaking, there is no time period before two hours when all the even coins are gone. After the two hour mark it cannot reasonably be said that even coins remain, since no distinct coin can have been left behind. However, since this condition only 'exists' instantly following the two hour mark, there has been no time to begin putting in or taking out the odd chips. Since the two hour mark ends the experiment, the even chips have been put in and taken out, but the odd chips remain untouched, never entering the dish.

That's my take on it anyway. At 2 hours and 1 second, where will the odd chips be? The 2 hour mark does not necessarily end the experiment, it's just that, before, nothing interesting happened after the 2 hour mark. If the majority are right, and there are 0 chips in the dish in the situation in the OP, then I think all of the odd chips must be instantaneously moved from not yet in the dish to inside the dish to taken out of the dish at the moment when t=2 hr.

William_Ashbless
12-11-2002, 04:35 PM
I think what is happening at the 2 hr mark precisely is somewhat ill-defined, but before, not all the evens have been in the dish, and after, all the evens are already out.

The problem is you cannot come up with a point in time where the odds have started moving through the dish! The description of the problem requires that the odd coins themselves have a point in time that they are removed from the dish. Saying the odd coins are taken from the dish after the 2 hour mark does no good, because the time at which a coin would be removed from the dish has no meaning (requires taking the logarithm of a negative number to solve). Before the 2 hour mark, only the evens are leaving the dish.

So as you say, they'd all have to leave the dish exactly at the two hour mark which is ill defined because the problem supposes every coin is moved at a different time (that is, after some time t since the last coin, a new coin is moved).

I claim this is enough to show that the odd coins never enter or leave the dish.

ZenBeam
12-11-2002, 04:44 PM
Tyrrell McAllister, thanks for your response. It gives me a chance to show the dilemma I'm seeing in something someone else wrote. I'm going to rewrite much of your post, applying it to my example 2, with the fly:

In the theorem that follows, I'm going to identify each move in the +X direction (hereafter "move" for brevity) with a number. So now N may be thought of as the set of all the moves.

Theorem:The set D of moves to the right not compensated for by an equal move in the -X direction (hereafter "antimove") at two hours is empty.
Proof: We show that each move is in Dc; the conclusion will then follow from the Lemma. In principle, Dc contains two kinds of moves:

(i)those moves that were never made before the two hour mark, and
(ii)those moves that are compensated for by an antimove before the two hour mark.

I think that everyone here agrees that the set of moves satisfying condition (i) is empty. That is, every move is eventually made at some time before the 2 hour mark. So the problem reduces to showing that every move in N is of type (ii).

To prove this, suppose that n is a move in N. Then n is compensated for by an antimove at time tn = 2 - (1/2)n-1(*). Note that, regardless of the value of n, we have that tn < 2. Therefore, move n is compensated for before the two hour mark, so move n is in Dc.

Since we showed this for an arbitrary move n in N, we've shown that every move in N is in Dc. Therefore, by the Lemma, D, the set of moves not compensated for by a corresponding antimove, is empty.

(ZenBeam again) We've just proved that in example 2, the fly ends up at the origin. How do you reconcile this with the fly approaching log(2) when the series in example 2 is summed? I reconcile this by asserting that this proof implicitly converts Eq. (1) to Eq. (2). This happens when each move is paired with its antimove in the sentence marked with (*). The sentence in your original post corresponding to the one with the (*) is likewise suspect. Certainly, if this example fails for a conditionally converging series, there is no reason to believe it will not also fail for a diverging series.

Now, you referred me to a previous post where you assert that "the more natural definition of H(infinity)" is zero, but that value of zero is based on reasoning like above, which gives an incorrect value for a case where we do know the value. At any rate, saying H(infinity) = 0 or the number of chips in the dish = 0 because I'm defining them to be zero is highly unsatisfying, and no better than saying the number of chips in the dish is infinite because I'm defining it to be infinite.

Cabbage
12-11-2002, 05:09 PM
knock knock, the experiment you're proposing can't be done with the standard division of time into "steps" that we've been using.

For example, in the original division, before each step occurs, only finitely many steps have already been taken. For your new experiment, some of the steps (the steps involving the odds) don't take place until infinitely many steps have already taken place.

Not that there's a problem with this--we only have to devise a new plan for the steps. The most obvious would be to simply repeat, in the following two hours, the same division into steps that we utilized in the first two hours. So essentially we have a two hour period involving the evens, followed by a two hour period involving the odds. It's essentially like doing the original experiment twice in succession, and now, at the end of the four full hours, the box will still be empty.

You may be interested in reading some on the ordinal numbers (http://mathworld.wolfram.com/OrdinalNumber.html). For example, in the original experiment, the steps were orderd by the ordinal omega. In our new experiment, the steps are ordered by the ordinal omega + omega.

Cabbage
12-11-2002, 05:54 PM
ZenBeam, I think you're missing Tyrrell McAllister's point--that we shouldn't be using limits in the first place.

Starting from the beginning, we're considering a time interval [0,2] (time expressed in hours).

We have a function f:[0,2] -> {0,1,2,3,...,omega (or infinite, if you prefer)}, that, when you plug in a given time, will spit out the number of chips in the box at that time.

The question we've all been considering is, what is f(2)?

Certainly, no one denies that (lim f(t) as t->2) = omega.

However, as any first year calculus student will tell you (provided they've been paying attention), you can't assume that

f(2) = lim f(t) as t -> 2.

This is true if and only if f is continuous at 2.

My (and others) approach to finding f(2) is simple. Simply analyze what chips are in the box at time 2, then count them. This is, after all, what f(2) is defined to be--how many chips there are in the box at time 2.

Your approach is more involved. You count the number chips before time 2, notice an upward trend, then assume this upward trend will be preserved, by a limiting process, at t=2.

This is your error. Can you demonstrate that f is continuous at t=2? If you can't, this approach must be discarded.

ZenBeam
12-11-2002, 09:02 PM
This is your error. Can you demonstrate that f is continuous at t=2? If you can't, this approach must be discarded.My "approach" has been to call into question your approach, and to ask you to show that your approach is valid. The reason I doubt your approach is that when I attempt to apply it to a problem where we do know the answer (my example 2 above), your approach gives the wrong answer. What you are doing is equivalent to starting with Eq. (1), and rearranging to get Eq. (2), and you haven't given any justification for doing so. Basically for the reason "we say so", the fly ends up at 0, when it had been approaching log(2).

Do you believe in example 2, that the fly is at X = 0 at t = 2? Does this make more sense to you than that the fly is at log(2)?

erislover
12-11-2002, 09:22 PM
Originally posted by ZenBeam
Equations are how you ensure you are removing chips in the right order. Well, in the first example, the equation forbids us from removing odd numbered chips. Thus, an infinite amount remain. In the second example, the equation compells us to remove every chip. Thus, none remain.

What am I missing?

Cabbage
12-11-2002, 09:36 PM
Do you believe in example 2, that the fly is at X = 0 at t = 2? Does this make more sense to you than that the fly is at log(2)?

Involving the fly seems to be a clever way of concealing the fact that you're still dealing with the limit of an infinite series. I haven't used any infinite series in concluding that the box is empty, so I fail to see how this calls my approach into question.

ZenBeam
12-12-2002, 07:39 AM
You keep avoiding answering those questions...

I fail to see how this calls my approach into question.I explained my reasoning in the paragraph above the one you quoted:The reason I doubt your approach is that when I attempt to apply it to a problem where we do know the answer (my example 2 above), your approach gives the wrong answer.Now I'm assuming here that your and Tyrrell McAllister's approaches are essentially the same.

Now, either you believe the fly is at zero at t=2 or you don't. The fly being at zero at t = 2 is, to me, counterintuitive. If you tell me you believe the fly is at zero at t=2, I'm going to ask the other mathmeticians who hang out here if they agree with that, and more to the point, if that is the consensus of current mathematical thinking. If you tell me you don't, I'm going to ask you why my rewrite of Tyrrell McAllister's proof fails. To emphasize, this is important because Tyrrell McAllister used essentially that same proof to show that the number of chips in the dish is empty, and your subsequent responses imply you agree with that proof.

knock knock
12-12-2002, 08:42 AM
Suppose god gets tired of dealing with an infinite number of chips. He just has one chip. He decides that this chip must always be labeled with one of the positive integers. At first he puts a "1" on it. After an hour, he erases the 1 and makes it a 2. A half hour later, he changes it to a 3. 15 minutes later, he changes it to a four. And so on. At any time before 2 hours, he has 1 chip with a positive integer on it, and as t-> 2 hours (from the left) that integer approaches infinity. Now, what does god have at the 2 hour mark? Everything from Tyrrell McAllister's proof seems to apply here, in slightly revised form. The chip always has a positive integer on it. After 2 hours, it cannot have a positive integer on it, because, for any integer n, I can tell you the moment when n was erased from it, and once it is erased, it never comes back. So there can be no positive integer on the chip. As long as the chip exists, it always has a positive integer on it (and can never stop having a positive integer on it, because whenever one is erased another replaces it instantaneously), it must not exist. He doesn't really have a chip.

That is just like the reasoning of those who say that there are no chips left after 2 hours in the case described in the OP. But how could the chip disappear just from writing and erasing numbers on it? Where would it go? The one chip must still be there, just like it always has been. It's just hard (or impossible) to say what it's labeled.

NE Texan
12-12-2002, 08:56 AM
I think that writing the sets out in some mathematical notation does help the problem a bit.

In owl's original problem:
The first set is "all integers" (except those divisible by 2), or:
S1 = {All n in N such that there is no m in N where n = 2m}

This will consist of positive integers n where {n = 2m + 1, and m is in N}. This will have the same cardinality as N, which is an infinite set (specifically, aleph null).

The second set is "all integers" (except those that have an integer higher than them), or:
S2 = {All n in N such that there is no m in N where m > n}

This will consist of positive integers where there is no higher integer - which is none of them, there is no highest integer. It's an empty set.

The problem comes when you try to count the chips by looking at what happens to a series of similarly constructed finite sets. That is what we are doing, if you think about it - after turn 1, you have one finite set; after turn 2, you have a different finite set; and so forth. It's tempting to expect the total "size" of S2 to be the limit of the finite sizes of these sets, but there is no mathematical guarantee that it will be. When the limit does not converge, we may say it "goes to infinity", but that does not mean that it converges at the cardinal number aleph null; it means that it does not converge, and we cannot extend the function in that way. I think that Cabbage and ZenBeam have explained the limits and their limits (snort) pretty well.

Cabbage
12-12-2002, 09:40 AM
knock knock, I would argue that the chip is blank at the end (the chip still exists, of course). Everything that was ever written on it has been erased, so it's blank.

ZenBeam, I'm not ignoring you, I just don't have time to respond at the moment.

knock knock
12-12-2002, 09:59 AM
Originally posted by Cabbage
knock knock, I would argue that the chip is blank at the end (the chip still exists, of course). Everything that was ever written on it has been erased, so it's blank.How could the chip be blank? Every time god erases a number from it, he simultaneously inscribes a new number on the other side.

Or, imagine a number system where each number n is represented by n dots. So 1 is . 2 is : 3 is :. and so on. For each move, God writes a new number on it - that is, he adds a dot. After infinitely many moves, there are infinitely many dots, right? Or are there somehow 0 dots?

knock knock
12-12-2002, 10:02 AM
I'm going to go back to the original problem and make a few changes in the procedure. I'll try not to make any important changes - that is, I'll try not to change any facts that have been used in the reasoning to support the majority opinion that no chips remain after 2 hours.

First of all, let's use gold coins instead of chips. 'Cause I like gold.

Now, have a move consist of removing a coin and then adding 2. Since these 2 steps occur together, perhaps instantaneously, we can do it this way instead of adding 2 and then removing 1. Every coin is still getting removed at the same time as before. Except we can't do this on the first move, so let's make the moves go as follows: 1) add 1 & 2, take away 1. 2) remove 2, add 3 & 4. 3) remove 3, add 5 & 6. And so on. Or, better yet, we can just start at the 1-hr mark, with only coin 2 in the dish, and begin with step 2).

Now, does it matter that God has this infinite pile of coins beforehand, or could he just mint them right before he puts them in? It doesn't matter, since we don't care exactly how he got them before he put them into the dish, or how long they existed beforehand. As long as they come in the order of the positive integers, our interest in the coins begins the moment they get put into the dish. So, we could imagine that God has an infinite vat of liquid gold, and on every move he reaches in and pinches, thereby creating 2 new coins, with the next 2 integers on them. Then he puts these into the dish.

What about coins that have already been taken out? Do they have to all go together into a pile? No, we don't care. They can be destroyed. God can create a coin right before it goes into the dish, and destroy it right after it comes out, and that has no effect on what's happening in the dish - each integer goes in, each integer comes out, according to a regular pattern that has a limit at t=2 hr.

Since God cares about the environment, he reduces the creation and destruction by recycling. Each coin that comes out can be destroyed by tossing it into the vat, where it melts, and that may at some point be used in the creation of a new coin. None of this effects the dish - each integer goes in, each integer comes out, even if the coin its on is made, in part, out of recycled gold.

Now, did any argument make use of the size of the coins, or of the fact that they were all the same size? Nope. So we can vary the size. We can make the coins progressively smaller. God doesn't want to use an infinite vat of gold where a finite will one suffice.

So God could make coin 2 out of 1 kg of gold, coins 3 & 4 out of 1/2 kg, coins 5, 6, 7, & 8 out of 1/4 kg, 9-16 out of 1/8 kg, and, in general, coins (2^n)+1 through 2^(n+1) out of 1/(2^n) kg of gold. This way, there is always 1 kg of gold in the dish. Since the coin that comes out is always the same weight as the 2 coins that go in, God could manage with exactly 1 kg of gold. The nth move would be as follows: God takes out the coin in the dish with the smallest label: n. He melts it down & forms it into 2 coins, half the weight of the original, identical to each other except that they are labeled with consecutive integers 2n-1 and 2n. He puts these new coins into the dish. Total elapsed time: as small as you want.

Now, imagine that the coins are stacked in a cylinder. They are all the same diameter, and only their thickness varies. Of course, they're all still labeled with integers. The nth move involves taking out a piece of the cylinder, cutting it in half, and removing the label on that piece and replacing it with a new label on each piece.

Since God strives for efficiency, he finds a fast way to do this. He doesn't even have to pick up the piece he removes. While leaving it in its place in the stack, he reaches around it with his hand, and uses his powers to transform it into the 2 newly labeled pieces. I don't know if he breaks it in half and relabels it, or if he melts it down and reforms it - it doesn't really matter. But he uses the same gold that was there.

Now, take a step back and look at the process. God starts with a solid 1-kg cylinder of gold, with a label on it (2). On each move, God leaves the cylinder where it is, but he breaks it in pieces and relabels the pieces. The whole cylinder is always there - it never moves - so it must be there after 2 hours. Where else would it go? Since the coins are getting thinner and thinner during the 2 hours, they must be infinitesimally thin by the 2 hour mark. Since they still have the same total height as at the beginning, there must be infinitely many of them. The numbers written on the pieces were increasing without bound, so I do not know how they are labeled after 2 hours. But that doesn't mean that they can't exist - that the whole cylinder suddenly disappears at the 2-hr mark, just because its pieces were being cut in half & relabeled up until then. So, in this case, there are infinitely many coins after 2 hours.

So, if you said that there were 0 chips after 2 hours in the original question, what happened? Are there still somehow 0 coins after 2 hours? If not, then which change in the process caused the answer to change?

ZenBeam
12-12-2002, 11:30 AM
Or, imagine a number system where each number n is represented by n dots. So 1 is . 2 is : 3 is :. and so on. For each move, God writes a new number on it - that is, he adds a dot. After infinitely many moves, there are infinitely many dots, right? Or are there somehow 0 dots?I really like this question.

ultrafilter
12-12-2002, 11:52 AM
The sum of an infinite amount of 1's (the dots in this case) is undefined.

RM Mentock
12-12-2002, 12:09 PM
Originally posted by knock knock
How could the chip be blank? Every time god erases a number from it, he simultaneously inscribes a new number on the other side.
Except when he finishes. :)

Or, imagine a number system where each number n is represented by n dots. So 1 is . 2 is : 3 is :. and so on. For each move, God writes a new number on it - that is, he adds a dot. After infinitely many moves, there are infinitely many dots, right? Or are there somehow 0 dots?
infinitely many dots, since he never erases any. I'm guessing it'll look a lot like his signature--well, that's what I would do.

knock knock
12-12-2002, 01:03 PM
Let's go back to that one chip with the dots. Let's leave that chip untouched, but add more chips to it. God starts with a one-dot chip. On the first move, he adds a dot to it. On the 2nd move, he adds a dot to it & adds another chip with 4 dots. On the 3rd move, he adds a dot to each of those chips & adds another chip with 6 dots. On the nth move, he adds a dot to each chip that is already there and adds a new chip with 2n dots. As a sequence of sets, this can be written: {2},{3,4},{4,5,6},{5,6,7,8},...,{n+1,...,2n},... , with each element in the nth set representing one chip in the dish after n moves, and the number of the element representing the number of dots on that chip. This is the same as the sequence of sets for the original procedure. In this case, though, we do not end up with 0 chips. We end up with infinitely many chips, each with infinitely many dots on them. Or, if we follow ultrafilter's statement, the number of chips at the end is undefined, and the number of dots on any given chip is undefined.

The sequence of sets for the 2 procedures are the same. So do the results have to be the same? Let's tweak this new procedure a little to make that question easier to answer (I hope). Instead of adding 1 dot to every chip that's already in the dish, on the nth move let's add n-1 dots to the chip with the smallest number of dots on it (n). Before, when going from {4,5,6} to {5,6,7,8}, old 4 became new 5, old 5 became new 6, old 6 became new 7, and 8 was brand new. Now, old 4 becomes new 7, old 5 & 6 stay 5 & 6, and 8 is brand new. This is a lot closer to the original procedure. And of course the timing remains the same as always.

Now, let's imagine that God takes chip #4 (or n) and adds 3 (or n-1) dots to it on the 4th (or nth) move. He also happens to have another chip available that is identical to this one - with exactly 7 (or 2n-1) dots on it. He takes his newly made 7 (or 2n-1) and his previously available 7 (or 2n-1) and holds one in each hand, trying to decide which to pick. He inspects them and finds them identical. Maybe he shakes them in his hands to mix them up & asks you which one to put into the dish. They look completely identical to you (though of course He knows their history), and so you pick one. Of course, all of this happens in no time flat.

Now, according to some arguments here, if you always pick the previously available chip, then the dish will end up empty. If you always pick the chip the he just took out and changed, then the dish will have infinitely many (or an undefined number of) chips in it at the end. How can the number of chips in the dish depend on which of 2 seemingly identical chips you choose to put in?

If you're thinking that the chip's history is somehow essential, I can make this example more magical and mysterious. God does not need to pick up and move chips - he is all powerful. Instead, he can merely hold the new chip (7 or 2n-1) in one hand, and point to the chip in the dish (4 or n). And ZAP, the next thing you know the chip in the dish has 7 dots on it (or 2n-1) and the one in his hand has only 4 (or n). And you ask God - what were you doing? Did you make the 2 chips switch places? Or, did you just add dots to the chip in the dish & remove dots from the chip in your hand, possibly transferring some of the dots on the chip in your hand to the one in the dish? And God says: "What's the difference?"

NE Texan
12-12-2002, 01:30 PM
Originally posted by knock knock
...imagine a number system where each number n is represented by n dots. So 1 is . 2 is : 3 is :. and so on. For each move, God writes a new number on it - that is, he adds a dot. After infinitely many moves, there are infinitely many dots, right? Or are there somehow 0 dots?

I was tempted to be a smartass and say that there is now a line segment on the coin - except that I know that's a higher cardinality than integers, so that's wrong. How about a Cantor dust?

RM Mentock
12-12-2002, 01:37 PM
Originally posted by NE Texan
I was tempted to be a smartass and say that there is now a line segment on the coin - except that I know that's a higher cardinality than integers, so that's wrong.
But the dots are finite, not infinitesimal, so it's OK.

Tyrrell McAllister
12-12-2002, 03:17 PM
Thank you, knock knock, for creating some interesting variations on the problem.

However, I think that an important change is made when we switch from decimal notation to dot-notation. Decimal notation, by definition, can only represent finite quantities, but dot-notation is much more powerful: it can be used to represent infinite quantities. In fact, dot-notation could be used to represent any infinite cardinal whatsoever. Therefore, this is a very non-trivial change to the parameters of the problem.

Case IA: "Suppose god gets tired of dealing with an infinite number of chips. He just has one chip. He decides that this chip must always be labeled with one of the positive integers. At first he puts a "1" on it. After an hour, he erases the 1 and makes it a 2. A half hour later, he changes it to a 3. 15 minutes later, he changes it to a four. And so on. At any time before 2 hours, he has 1 chip with a positive integer on it, and as t-> 2 hours (from the left) that integer approaches infinity. Now, what does god have at the 2 hour mark?"

My Answer: At the 2 hour mark, God has a blank chip. When Cabbage gave this answer, you responded, "How could the chip be blank? Every time god erases a number from it, he simultaneously inscribes a new number on the other side." Ah, but God can write an integer on the chip only if there are any integers left to write. But at the two hour mark, he's run out of integers, and so can write nothing. Moreover, he's erased everything he ever wrote. Thus the chip is blank.

Case IB: "Or, imagine a number system where each number n is represented by n dots. So 1 is . 2 is : 3 is :. and so on. For each move, God writes a new number on it - that is, he adds a dot. After infinitely many moves, there are infinitely many dots, right? Or are there somehow 0 dots?"

My Answer: There are infinitely many dots on the chip. Unlike in the case were God was restricted to writing integers, He now has something to write at the two-hour mark: infinitely many dots.

Case II: "He doesn't even have to pick up the piece he removes. While leaving [a coin] in its place in the stack, he reaches around it with his hand, and uses his powers to transform it into the 2 newly labeled pieces. I don't know if he breaks it in half and relabels it, or if he melts it down and reforms it - it doesn't really matter. But he uses the same gold that was there. . . . God starts with a solid 1-kg cylinder of gold, with a label on it (2). On each move, God leaves the cylinder where it is, but he breaks it in pieces and relabels the pieces."

My Answer: God has 1-kg of gold, broken into infinitely many peices, each of which is 0-kg. In fact, if I'm not mistaken, he has uncountably many pieces. Can God write an integer on such insubstantial films of nothingness? If we allow that he can, then I'd say that they are all blank because he ran out of integers to write on them. Certainly, almost all of them are blank, because there are uncountably many pieces and only countably many integers.

Case IIIA: "Let's go back to that one chip with the dots. Let's leave that chip untouched, but add more chips to it. God starts with a one-dot chip. On the first move, he adds a dot to it. On the 2nd move, he adds a dot to it & adds another chip with 4 dots. On the 3rd move, he adds a dot to each of those chips & adds another chip with 6 dots. On the nth move, he adds a dot to each chip that is already there and adds a new chip with 2n dots. As a sequence of sets, this can be written: {2},{3,4},{4,5,6},{5,6,7,8},...,{n+1,...,2n},... , with each element in the nth set representing one chip in the dish after n moves, and the number of the element representing the number of dots on that chip."

My Answer: I agree with your answer. We end up with infinitely many chips, each with infinitely many dots on it.

Case IIIB: "Let's tweak this new procedure a little to make that question easier to answer (I hope). Instead of adding 1 dot to every chip that's already in the dish, on the nth move let's add n-1 dots to the chip with the smallest number of dots on it (n). Before, when going from {4,5,6} to {5,6,7,8}, old 4 became new 5, old 5 became new 6, old 6 became new 7, and 8 was brand new. Now, old 4 becomes new 7, old 5 & 6 stay 5 & 6, and 8 is brand new."

My Answer: Again, we end up with infinitely many chips, each with infinitely many dots on it.

Cases IIIC and IIID: "Now, let's imagine that God takes chip #4 (or n) and adds 3 (or n-1) dots to it on the 4th (or nth) move. He also happens to have another chip available that is identical to this one - with exactly 7 (or 2n-1) dots on it. He takes his newly made 7 (or 2n-1) and his previously available 7 (or 2n-1) and holds one in each hand, trying to decide which to pick. He inspects them and finds them identical. Maybe he shakes them in his hands to mix them up & asks you which one to put into the dish. They look completely identical to you (though of course He knows their history), and so you pick one. Of course, all of this happens in no time flat."

"I can make this example more magical and mysterious. God does not need to pick up and move chips - he is all powerful. Instead, he can merely hold the new chip (7 or 2n-1) in one hand, and point to the chip in the dish (4 or n). And ZAP, the next thing you know the chip in the dish has 7 dots on it (or 2n-1) and the one in his hand has only 4 (or n). And you ask God - what were you doing? Did you make the 2 chips switch places? Or, did you just add dots to the chip in the dish & remove dots from the chip in your hand, possibly transferring some of the dots on the chip in your hand to the one in the dish? And God says: "What's the difference?"

My (non)Answer: I'm going to have to think more about these when I have the time. Right now, my inclination is to say that, God's protestations to the contrary, it really does make a difference which it is that God is doing. See this (http://boards.straightdope.com/sdmb/showthread.php?postid=2698634#post2698634) post of mine from the previous page where I gave some of my thoughts on a related phenomenon.

Chronos
12-12-2002, 03:24 PM
I really like knock knock's analysis of the cylinder of gold, because we don't need to be God to pull that one off. Let's say that, instead of a physical cylinder of gold, we just use the line segment [0,1), and we're dividing it into smaller and smaller line segments (for sake of definiteness, let's consider all of the subsegments to be half-open, so they can exactly add up to the original segment). I can define a labelling scheme as a function of time which exactly simulates the gold cylinder situation, and just say that that's the labelling scheme. But if I do write down that labelling scheme, that doesn't mean that two hours later, the number line will all of a sudden have to skip directly from 0 to 1. The line segment is still there. The manner in which it's divided up is not well-defined after two hours, but the segment is there nonetheless.

panamajack
12-12-2002, 04:17 PM
With due respect to Tyrell McAllister's excellent analysis so far, I'll add my comments on this part :

originally posted by knock knock
Now, according to some arguments here, if you always pick the previously available chip, then the dish will end up empty. If you always pick the chip the he just took out and changed, then the dish will have infinitely many (or an undefined number of) chips in it at the end. How can the number of chips in the dish depend on which of 2 seemingly identical chips you choose to put in?

Calling them 'seemingly identical' is only circular reasoning, though. There is a distinction between the chips -- those that have been in the bowl and those that haven't been in the bowl.

If IIIC = use the previous (nth) chip and IIID = use the new (2n-1) chip, in IIIC a chip put in the bowl is never removed; in IIID it is. Suppose we paint the #4 chip black. Now looking at the state of the bowl after each step, we have :

IIIC IIID
1 1
2 2
3 4 3 4
4 5 6 4 5 6
5 6 7 8 5 6 7 8
6 7 8 9 10 6 7 8 9 10
7 8 9 10 11 12 7 8 9 10 11
8 9 10 11 12 13 14 8 9 10 11 12 13 14
etc. etc.

In IIIC, at no time does the black chip ever leave the bowl. It gets marked with some higher (odd) number. In fact, had we painted all the (originally even) chips black, we end up with a bowl full of black chips in IIIC. In IIID, the chips actually leave the bowl. So those cases are not identical.

Now, if God's just 'zapping' the chips, everything's fine up until t=2. But what actually happens at infinity? Following the 'blank coin' example, there cannot be a 'infinity' coin to zap with. I'll admit this doesn't explain much, but I have difficulty trying to explain how one 'accomplishes' an infinite number of tasks.

Cabbage
12-12-2002, 07:56 PM
ZenBeam, I'm sorry, it's the end of the semester, and all of a sudden I've got a big pile of work to do. I'll try and make a response to you now, but it's going to be much briefer than I'd like.

Actually, I haven't addressed your infinite series example specifically because I'm not sure you wrote what you intended to write. We know

log(2) = 1 - 1/2 + 1/3 - 1/4 +....

but that's not what I'm seeing in your second example, which reads to me as
1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 + ...

I don't know offhand what that series converges to (if it converges at all).

I figure that's probably irrelevant to your point, so I was trying to address your point specifically, which seems (to me) to be summed up in two statements:
1. The position of the fly always (even at the end) corresponds to the number of chips in the box (or maybe the height of the chips).
2. The movement of the fly is continuous.

None of this changes my position. If I grant you 1., I simply can't grant you 2. If I grant you 2., I simply can't grant you 1. The fly is merely an additional element to the experiment--maybe it moves continuously, maybe its position always corresponds to the height of the chips, but I claim it can't do both.

My apologies if I've actually missed your point entirely.

knock knock, I applaud you for your interesting variations on the problem.

Tyrrell McAllister, I applaud you for your analysis of knock knock's variations.

I wish I could respond more, but I gotta get back to work. Hopefully in a day or two I'll be able to participate more.

ZenBeam
12-12-2002, 09:27 PM
Try this:
log(2) = 1 + (1/2 - 1) + 1/3 + (1/4 - 1/2) + 1/5 + (1/6 - 1/3) + ...

Anyway, my point remains that I'm trying to use the same methods Tyrrell McAllister used to determine the dish is empty on a different problem, that of the fly in example 2. Forget about the chip thicknesses. I'm using Tyrrell McAllister's proof because he's the only one who attempted to actually show it with better than handwaving arguments.

Can anyone (and I'm especially looking at the "dish is empty" crowd, although those of you with outside lives get a bit of a pass) confirm or deny whether I've correctly modified Tyrrell McAllister's proof in this post (http://boards.straightdope.com/sdmb/showthread.php?s=&threadid=148106&postid=2704534#post2704534) to show that the fly is at zero for t=2? Is the fly at 0 or at log(2) at t=2? Surely, if everyone can answer the OP, whcih doesn't even converge, someone can figure this one out.

While we're at it, what about Tyrrell McAllister's original proof showing the dish is empty, about three posts prior to the linked one. Is this a valid proof in the eyes of the rest of the DIE gang?

erislover
12-12-2002, 10:04 PM
In the case where we remove the largest, the sum of all the numbers on the chips in the bowlwhatever those numbers areat time k is
Sn=0k (2n+1) = (k+1)2

In the case where we remove the smallest, the sum of all the numbers on the chips in the bowlwhatever those numbers areat time k is... well, I can't develop a series for it as the terms change each time n changes, but there is a recursive sequence of partial sums, which is
kn+1 = kn + 5 + 3(n-1); k0=2

If the sum of the numbers on the chips grows without bound, then I don't think one can claim the bowl is ever empty, whether we can name a single number or not.

I'm so confused.

But if we cannot create a series representation of the second case, I'm not sure I can say my argument holds any water.

knock knock
12-12-2002, 10:29 PM
Thanks everyone for your compliments, and for your interesting arguments. I'm mostly going to be responding to Tyrrell McAllister's post here, and, since I don't want to make my post longer than it already is with an unwieldy amount of cutting and pasting, I'll just refer to the cases as they have been nicely labeled above and give a few important quotes.

Case IA: "Ah, but God can write an integer on the chip only if there are any integers left to write. But at the two hour mark, he's run out of integers, and so can write nothing. Moreover, he's erased everything he ever wrote. Thus the chip is blank." I don't see how this "running out" of things to write can happen any sooner than a running out of things to erase. Every time God is in the process of erasing an integer n, he can always write another integer, namely n+1. Imagine that God had this quirk - he abhors a blank chip more than nature abhors a vacuum. So every time he erases a number from a chip, he is always sure to write another number on the chip simultaneously, if not sooner. Now, would this keep God from completing the task I described? I think He'd still be able to do it just fine, because this is precisely what he does on every move - he replaces a number with a higher one. And there's always exactly one number written on the chip.

Case II: (the cylinder) "God has 1-kg of gold, broken into infinitely many peices, each of which is 0-kg. In fact, if I'm not mistaken, he has uncountably many pieces." In this case I believe you are mistaken, possibly because I wasn't quite clear to begin with. On each move, God only breaks one of the pieces of the cylinder in half (and it's always one of the largest remaining pieces of the cylinder). I was trying to keep this problem isomorphic with the original. The number of pieces of cylinder after n moves = number of dots on coin after n moves = number of chips in dish after n moves. The number of pieces increases by 1 each move, so it stays countable.

"I'd say that they are all blank because he ran out of integers to write on them." I'm not sure if this comment is based on the belief that there are uncountably many pieces. If he used dot notation, would they still be blank?

You agree that the cylinder is there, and that its mass is still 1 kg., though it's composed of infinitely many "insubstantial films of nothingness." Now, topple the cylinder over. Knock it into an amorphous pile (but don't break or combine anything). Now, maybe you can't see it, but, if the whole thing is sitting on a scale, the scale should still read 1 kg.

And this would still happen, presumably, if each move consisted of taking out a coin (i.e. piece of cylinder), melting it down, and smelting 2 new coins. Although I guess you would say these coins wouldn't really be "new". I gather that you would say that the crucial difference between the different conditions I stated in that post (http://boards.straightdope.com/sdmb/showthread.php?threadid=148106&perpage=50&postid=2707984#post2707984) is whether the coin going into the dish is made out of the material that was taken out, or if it was made out of new material.

Case IIIC,D: And here's the fun one. I'm not sure, panamajack, if you're allowed to paint a chip black. The 2 chips you choose between are supposed to be indistinguishable except for their history. Tyrrell McAllister's historical view is intriguing, but here's an argument for why history doesn't matter:

Say God is on an arbitrary one of his moves (call it the nth) and He has kindly stopped time to ask you the question: which chip should go in the dish? Now, the chips look identical to you, and you are feeling very nervous and concerned about your choice. You have a bet with your friend who says that there won't be any chips in the dish at the end of this. You bet that there will be chips, and you're suspecting infinitely many of them. You're feeling a lot of pressure because you are familiar with arguments that show that the result depends entirely on your choices (and also because you're telling God what to do, but you're already kinda getting used to that). But then you realize: after this move, there will be n chips in the dish, whatever choice I make. So I can start from there, with the problem beginning with n chips. Those n chips can be the special, marked chips that I worry about. My choice here is irrelevant.

And you're right. You can choose whichever chip on the first n moves, and then choose right (i.e. as in Case IIIC) the rest of the way, and the number of chips will end up infinite.

So, on any move n, your choice is irrelevant. Since n was an arbitrary move, your choice on any move is irrelevant. So all of your choices are irrelevant.

And I just realized that my argument began with the statement "here's an argument for why history doesn't matter:" and ended "So all of your choices are irrelevant." I could change that, but instead I'll just bring it to everyone's attention.

panamajack
12-13-2002, 02:49 AM
I was painting it black only so we could follow it; to see the difference between a chip always chosen to remain in the bowl and the one that gets removed. I hope that shows that if you do care about which chip is where, it makes a difference. I'll admit to not having a completely solid answer to the 'zapping' or choosing case.

But it did make me think of an interesting and potentially annoying variation :

Starting similarly to the original problem statements ...

Suppose that after adding two chips to the bowl, God now flips a perfect coin (or rolls a three-sided die and ignores the 3's) to decide whether he will remove the largest or the smallest chip from the pot. What is the expected number of chips left in the bowl after 2 hours?

It's too late for me to think about the answer to that one, so I'll flip a coin and say it's infinite.

ZenBeam
12-13-2002, 11:31 AM
I have a modification of example 2 (http://boards.straightdope.com/sdmb/showthread.php?s=&postid=2703341#post2703341) with the fly, I'll call it example 2b. (In example 2, ignore anything about chips, I think that just confused people.) I'm going to use this to show a contradiction in Tyrrell McAllister's method of proof.

For both this example and example 2, I need to further specify that the fly only travel in the last, say, 1/60th of the time period. So in the first step, the fly rests 59 minutes, then travels in the final minute before t=1hour, in the next step it rests 29.5 minutes and travels in the final 30 seconds before t=1.5 hours, and so forth. Further, the fly travels at a uniform speed during the travel period. The fly's motion in both examples is now fully specified for t<2.

I need an additional equation for reference.

Eq. (4):
S4 = alpha + b0 - c0 - d0 - e1 + a1 + b1 - c1 - d1 - e2 + a2 + ...

Example 2b.
In this example, the fly travels in the X direction in the following amounts:
1 + 1/2 - 1/2 - 1/6 - 1/3 + 1/3 + 1/4 - 1/4 - 1/20 - 1/5 + 1/5 + 1/6 + ...

so alpha and all of ai, bi, ..., ei are positive.

Once again, I'm going to rewrite much of Tyrrell McAllister's proof (http://boards.straightdope.com/sdmb/showthread.php?s=&postid=2703767#post2703767) showing the dish is empty, applying it now to my example 2b, with the fly:

In the theorem that follows, I'm going to identify each move in the +X direction (hereafter "move" for brevity) or in the -X direction (hereafter "antimove") with a number. So now N may be thought of as the set of all the moves and antimoves. alpha is move 1, b0 is move 2, c0 is antimove 3 and so forth.

Theorem: The set D of moves not compensated for by an equal antimove and antimoves not compensating a move at two hours is the set consisting of move alpha, and all antimoves di.
Proof: We show that each move Or antimove) ai, bi, ci, ei is in Dc and that alpha and di are in D. In principle, Dc contains three kinds of moves and antimoves:

(i)those moves or antimoves that were never made before the two hour mark,
(ii)those moves that are compensated for by an antimove before the two hour mark,
(iii)those antimoves that are compensating a move before the two hour mark,

while D contains:

(iv)those moves that were never compensated for by an antimove before the two hour mark, and
(v)those antimoves that are never used to compensate a move before the two hour mark.

I think that everyone here agrees that the set of moves and antimoves satisfying condition (i) is empty. That is, every move or antimove is eventually made at some time before the 2 hour mark. The set of every move bi satisfies condition (ii) (compensated by antimove ci).

To prove this, suppose that n is a move bi in N. Then n is compensated for by the antimove n+1 (ci) immediately following. Note that, regardless of the value of n, we have that tn < 2. Therefore, move n is compensated for before the two hour mark, so move n is in Dc.

Likewise the set of every move ai also satisfies condition (ii) (compensated by ei). Each of the set of antimoves ci and ei satisfies conditon (iii) (compensating bi and ai respectively).

alpha clearly satisfies condition (iv) (since all antimoves are less than 1). The set of moves di satisfy condition (v) (all the moves except alpha are already compensated by an antimove, and alpha is too large). Thus the theorme is proved.

The set of moves and antimoves satisfying conditons (i), (ii), and (iii) do not contribute to the fly's final position (they were either never made, were compensated for, or were compensating a move). The remaining moves and antimoves satisfying (iv) and (v) can now be easily summed, since they form an absolutely convergent sequence:

1 - 1/6 - 1/20 - 1/42 - ...

The sum converges (absolutely) to log(2). So for example 2b, the fly ends up at log(2).

I've already shown, using the same method, that in example 2, the fly ends up at zero (http://boards.straightdope.com/sdmb/showthread.php?s=&postid=2704534#post2704534).

The important point here is that the fly in example 2, and the fly in example 2b is the exact same fly. The fly in example 2 cannot. e.g travel 1 in the -X direction (i.e. travel -1) without traveling 1/2 then 1/6 then 1/3, all in the -X direction. Likewise, the fly in example 2b cannot travel 1/2 then 1/6 then 1/3, all in the -X direction without having traveled a total of 1 in the -X direction. Examples 2 and 2b are the same example.

Thus, I've used Tyrrell McAllister's method of showing the dish is empty on a different problem, and obtained two different answers. Tyrrell McAllister's method of showing that the dish is empty is flawed.

I'm not going to track down every statement that the dish is empty, since they're all variations on the same theme, and the same approach can be taken onal of them. For example, the tersest "proof" that the dish is empty is this one, for which I'll give an equally terse refutation:

My (and others) approach to finding f(2) is simple. Simply analyze what chips are in the box at time 2, then count them. I can simply analyze what moves are not compensated for in examples 2 and 2b and sum them up, but when I do so I get zero in the one case, and log(2) in the other, even though examples 2 and 2b are the same problem.

I therefore feel justified in stating that every proof offered in this thread that the dish is empty is flawed. Therefore no valid justification has been given for asserting that the dish is empty. I'll further assert that any future "proof" offered that the dish is empty needs to be applied to examples 2 and 2b, showing that different answers to these two examples are not possible. This has been a lot of work, and I'm not going to repeat it for every different phrasing that comes along.

Cabbage
12-13-2002, 12:49 PM
I've still got a lot of work to do, so I'll have to be brief, but:
I can simply analyze what moves are not compensated for in examples 2 and 2b and sum them up, but when I do so I get zero in the one case, and log(2) in the other, even though examples 2 and 2b are the same problem.
The problem is, I'm dealing with a set, while you're dealing with an infinite series.

With an infinite series, if I rearrange the terms of the series, I can end up with it converging to different values. With a set, if I rearrange the elements of the set, I still have the same set. You can't find fault with my argument by considering series, because my argument has nothing to do with series in the first place.

Also, you're still making the assumption of continuity, by implicitly assuming that the fly moves continuously (i.e., that the fly converging on the limit of the series of its moves).

As far as I'm concerned, you are free to assume the fly moves continuously, but then you'll have a hard time convincing me that it's in any way analagous to the original problem involving the number of chips in the dish.

Cabbage
12-13-2002, 12:55 PM
Oh, and why won't anyone respond to my variation, mentioned earlier?:

In the original problem, God throws a few positive integers in at a time.

In my variation, God decides to just throw them all in at the beginning, rather than merely two at a time. In other words, the dish already contains all the positive integers at the start. Then he begins removing them, removing the smallest one at each step.

How many chips are left in the dish when he's finished?

ZenBeam
12-13-2002, 02:17 PM
The problem is, I'm dealing with a set, while you're dealing with an infinite series.In both my proofs, I'm dealing with a set of moves. Only at the end, when the set has been reduced to a set of moves (possibly empty) which form an absolutely convergent series do I sum the series.

Also, you're still making the assumption of continuity, by implicitly assuming that the fly moves continuously (i.e., that the fly converging on the limit of the series of its moves).I specifed that the fly moves continuously and still got two different answers. And neither you nor anyone else has ever answered where the fly is at t=2. If the number of disks at t=2 is answerable, surely the fly's location is also.

erislover
12-13-2002, 02:23 PM
Originally posted by Cabbage
Oh, and why won't anyone respond to my variation, mentioned earlier?:

In the original problem, God throws a few positive integers in at a time.

In my variation, God decides to just throw them all in at the beginning, rather than merely two at a time. In other words, the dish already contains all the positive integers at the start. Then he begins removing them, removing the smallest one at each step. Yeah, that's why I liked to rephrase it as "God removed all even numbers" and "god removes all numbers". When it is phrased like that it does seem to be obvious.

I find both arguments totally convincing. And I like recursive sequences.

Cabbage
12-13-2002, 07:39 PM
ZenBeam, let me check that I'm understanding you. Consider one of your earlier series:

1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 + ...

As I understand it, you're offering, in some sense, two different interpretations of the terms in the above series:

1. +x and -x correspond to the fly moving x units to the right (move) and x units to the left (antimove), respectively.

2. (Relating it to our original example) +1/n and -1/n correspond to adding chip n to the box, and removing chip n from the box, respectively.

Assume the fly moves continuously, so that the location of the fly at t=2 will be the sum of the infinite series.

My original argument was that, according to the series, every chip is eventually removed, since -1/n occurs somewhere in the series.

Your argument is that this doesn't make sense, since if we simply rearrange the series, we can get the fly's motion to converge at various points.

I claim that one has nothing to do with the other. Sure, we can rearrange the series and get the fly going to different places, but, no matter how we rearrange the series, I still claim the dish is empty--I see no contradiction here. This is, of course, assuming that our rearrangement makes sense. For example, suppose our rearrangement begins:

-1/2 + 1/2 +....

This is not a proper rearrangement; it doesn't make sense to remove chip 2 before we place chip 2 in the dish, so we have to discard arrangments such as these (where -1/n comes before +1/n) on the grounds that it doesn't preserve the nature of the original problem.

Also, we must not allow discard combining terms. For example, the original series

1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 + ...

is significantly different from

1 + 1/2 - 1 + 1/3 - 1/4 + 1/5 + 1/6 - 1/3 + ...

since clearly, when it comes to discussing the dish, "adding chip 4 and removing chip 2" is not equivalent to simply "removing chip 4".

Again, my claim is that, regardless of what the fly may do, any rearrangement (subject to the above restrictions) will leave the dish empty of chips.

Let me stop it here and ask if I've understood you so far; please correct me if I haven't, and we'll take it from there....

knock knock
12-13-2002, 10:41 PM
Does anyone want to respond to the argument I made in this post (http://boards.straightdope.com/sdmb/showthread.php?threadid=148106&perpage=50&postid=2711857#post2711857) for Case IIIC,D? That was a situation like in the OP, except, whenever god was supposed to put in 2 chips (say, the 7 & 8) and take out 1 (the 4), he puts the 8 in, takes out the 4, relabels the 4 to be a 7, and asks you whether to put this relabeled 7 or a fresh 7 into the dish. Though the 2 chips look identical, Tyrrell McAllister suggested that, if you always choose the chip that had been in the dish, you will end up with infinitely many chips, while if you always choose the new chip, the dish will end up empty. Earlier (http://boards.straightdope.com/sdmb/showthread.php?threadid=148106&perpage=50&postid=2698634#post2698634), he brought up the idea of a chip being "tagged" for the whole process, so that identical-looking chips aren't really identical if they have different histories. My argument against this historical tagging went as follows:

Consider any arbitrary move n. Whether you choose the chip that was in the dish or the new chip, there will be n chips in the dish after this move. As long as there are a finite number of chips in the dish, you do not need to worry about which choice you make, since you can make the number of chips in the dish end up infinite by always picking those chips that are in the dish after this move. So, whether the old chip goes back into the dish or the new chip goes in on this move is irrelevant. Since the move n is arbitrary, the choice of chip is irrelevant on any move. So, it makes no difference if you choose the old chip on every move or if you choose the new chip on every move. Since the number of chips at the end is infinite if you always choose the old chip, and whether you choose the old chip or new chip is irrelevant, the number of chips at the end is infinite if you always choose the new chip. And that is precisely the situation raised in the OP.

Originally posted by Cabbage
Oh, and why won't anyone respond to my variation, mentioned earlier?:

In the original problem, God throws a few positive integers in at a time.

In my variation, God decides to just throw them all in at the beginning, rather than merely two at a time. In other words, the dish already contains all the positive integers at the start. Then he begins removing them, removing the smallest one at each step.

How many chips are left in the dish when he's finished?I'm not sure if this set is well-defined, but if it is, it looks like it should be empty.

The point of this, I assume, is that the set we're interested in is always a subset of this set. That is true after any finite number of moves, but it doesn't necessarily hold at the end. I could always make a discontinuity claim like you've been making against ZenBeam. The finite sets in case IIIC (where each move consists of adding a chip & relabeling a chip) are also subsets of the sets in this sequence, but it seems that everyone agrees that in IIIC we end up with a dish with infinitely many chips in it.

To erislover - we can't just say that "god removes all numbers", because that is what's being argued. After the nth move, God has all the chips up to 2n, except he's removed the bottom half (in the other case, he'd removed all the even numbers). Once the number of chips we're dealing with becomes infinite, it doesn't make sense to say that he's removed the bottom half, so it's hard to know what's there & how many are there. (In the other case, it still made sense to say that he'd removed all the even numbers.)

12-13-2002, 11:31 PM
OK, I have a few variations, and I want to see how this works for them.

Case One:

God creates two chips, identical in every aspect, and puts them in a very special bowl. The bowl instantaneously destroys one of the chips that enter it, every time. These events take place in zero time, and God schedules these events at the sequentially reduced intervals that will accomplish an infinite number of creation events in two hours.

How many chips?

Case Two:

God creates numbered chips, two sequential integers beginning with 1 and 2. God never repeats a number. The bowl always destroys the lower numbered chip that enters at each step. The devil is hiding in the bowl. Each time a new chip enters the bowl, he changes all the other chips by making each one show the next integer to the one currently on it. The devil can never destroy chips; he only changes numbers. Again, all these events take place in zero time, and the events are scheduled as before.

First God throws in 1 and 2. There are no chips in the bowl, so the devil has no chips to change. The bowl destroys number one, and number two is in the bowl. God throws in number 3 and 4. The bowl destroys chip number 3, and the devil turns the number 2 into a 3.

Now how many chips are there after two hours?

Are the chips still numbered sequentially? If they are not, why not? And if they are, isn't one of them the lowest value integer in the bowl, despite the fact that all the chips in the bowl are numbered with integers which are infinite in value?

How is that different from the original problem?

Tris

Cabbage
12-14-2002, 06:28 PM
knock knock
I'm not sure if this set is well-defined, but if it is, it looks like it should be empty.
Why wouldn't it be well defined? All that's really happening here is the principle of induction on the natural numbers:

At the end, clearly 0 is not in the dish, since it's removed in the first step. If n is not in the dish, it was removed at some step. After this step was taken, n+1 was the smallest chip in the dish, hence it was removed in the following step, implying n+1 is also not in the dish. By induction, no natural numbers are in the dish, and since only natural numbers are in the dish, the dish must be empty.

Are you saying something is wrong with induction?
The point of this, I assume, is that the set we're interested in is always a subset of this set. That is true after any finite number of moves, but it doesn't necessarily hold at the end.
Well, I just proved they're both empty, so we won't have to worry about that any more.

Finally, about your variations. Admittedly, I haven't had time to look at each of them in detail (honestly, some not at all), but what seems to be going on in the ones I did see was a confusion over "chips" versus "natural numbers".

In the original problem, there was a bijection between chips and numbers--every chip had exactly one number written on it, and every number was written on exactly one chip.

This is not true for your examples (or yours, Triskadecamus). Chips are getting their numbers erased, new numbers are being put on them, and so forth. We no longer have an direct correspondence between chips and numbers. We can talk about chips, and we can talk about numbers, but what's true of one may not be necessarily true of the other in these examples.

With this in mind, is there a particular example of yours I should consider? (I still have a lot of work to do at the moment, I only come here when I take a break and don't have time to review all your examples, so if you could point me to something specific, I'd appreciate it).

knock knock
12-15-2002, 12:50 AM
Cabbage - I'll assume that the set in your example is empty. When I said "The point of this, I assume, is that the set we're interested in is always a subset of this set. That is true after any finite number of moves, but it doesn't necessarily hold at the end." "the set we're interested in" refers to the set in the OP. Is that what you're referring to when you say you proved that they're both empty? I agree that the set of chips from n+1 to 2n is a subset of the set from n+1 to infinity, and I think the set from n+1 to infinity goes to the null set as n goes to infinity, but that does not necessarily mean that the set from n+1 to 2n goes to the null set. There could be a discontinuity there.

The most important examples, I think, are IIIC and IIID (Tris's example looks a lot like IIIC). You can look at my last post for an explanation this - in one case, there is renumbering, and in one case, there is not. I tried to argue that the case with renumbering (IIIC) ends up with infinitely many chips, and that the case without renumbering (i.e. with only replacement of chips - IIID - which is identical to the OP) must end up the same.

Cabbage
12-15-2002, 02:37 AM
I agree that the set of chips from n+1 to 2n is a subset of the set from n+1 to infinity, and I think the set from n+1 to infinity goes to the null set as n goes to infinity, but that does not necessarily mean that the set from n+1 to 2n goes to the null set. There could be a discontinuity there.
But the exact same set of chips are either in the dish (my example) or will be put in the dish (original example) in each case. To say that some are left in the original example but not mine is to say that some chips are removed in my example which were not removed in the original example, which is clearly not the case.

The most important examples, I think, are IIIC and IIID (Tris's example looks a lot like IIIC).
Let me go back to considering my example and the original example first, then I'll get to yours.

As I mentioned earlier, one of the things that makes the original example special is that we have a direct correspondence between chips and numbers, so we can speak of them interchangeably. This is also true of my example. And in each of these examples, we can reason that, since all natural numbers have been taken out, all chips must have been taken out.

Let's try a variation of my example (starting with all natural numbers in the dish and removing the smallest at each step). I'll adopt your idea of the number being written on the chip in the form of whatever number of dots.

Now suppose that at each step, instead of removing the smallest chip, God simply adds a single dot to each chip. Let's consider what happened--clearly no chip has been removed, yet clearly a number has been removed (the smallest number represented on a chip is no longer represented by any chip).

We've now lost the chip<->number correspondence, and must consider each individually--what happens to the chips may not happen to the numbers, and vice versa.

For the numbers: All the natural numbers will be removed from the dish, since, at some point, each natural number will be the smallest one left, hence it will be removed.

For the chips: No chips are ever leaving the dish. At the end, we'll still have an infinite number of chips in the dish.

And each of the infinitely many chips will have infinitely many dots on them. No integers are left in the box, yet the chips remain.

Now, similar to your example, suppose that as each step is taken, you don't know what God is doing. The dish goes from containing {3,4,5,...} to containing {4,5,6,...}, but you don't know if chip 3 was removed, or if one dot was added to every chip.

In either case, certainly all the integers will be removed. However, it's impossible to determine what will happen to the chips at the end, since you don't have knowledge of whether we have a one-to-one correspondence between chips and numbers.

I think it should be clear now what I would argue in your examples IIIC and IIID. If God is replacing the chips (and not relabeling them) we again have the chip<->number correspondence. All natural numbers have been taken out, therefore all chips have been taken out.

On the other hand, if God is relabeling the chips (adding dots on them), all the natural numbers are still being taken out, but we end up with infinitely many chips, each with infinitely many dots on them.

And if we don't know which God is doing, we don't have enough information to conclude what will happen in the end to the chips (though in either case we know all natural numbers are removed, so we still know that).

ZenBeam
12-15-2002, 09:20 PM
ZenBeam, let me check that I'm understanding you.You are focusing here on example 2 only. By itself, example 2 is a weaker argument than when it is combined with example 2B.
Assume the fly moves continuously, so that the location of the fly at t=2 will be the sum of the infinite series.I assume the fly moves continuously for t<2. In neither example 2 or 2b have I needed to state that whether the fly's motion is continuous for t<=2. Before I had example 2B, the answer that the fly is at 0 at t=2 was merely (one could argue) counterintuitive. With the addition of example 2B, I have a contradiction.

Your argument is that this doesn't make sense, since if we simply rearrange the series, we can get the fly's motion to converge at various points.While I agree with this statement, I'm following Tyrrell McAllister's proof without rearranging the series. My argument is not simply that rearranging the series can get the series to converge at different points. My argument is that the same scenario, merely by virtue of being described differently, yields two different answers. The fly ends up at two different locations at t=2.

This occurred to me reading your last post. Maybe its been answered previously, but it's late and I'm tired:
On the other hand, if God is relabeling the chips (adding dots on them), all the natural numbers are still being taken out, but we end up with infinitely many chips, each with infinitely many dots on them.What if God always writes an integer on each chip, in addition to the dots, and equal to the number of dots (writing the new number, then erasing the old if he adds a dot)? What numbers are written on those chips with infinitely many dots?

Cabbage
12-15-2002, 11:39 PM
ZenBeam, really, the very simple answer is that Tyrrell McAllister's applies to sets, not the sum of an infinite series. I've skimmed over your arguments, but it's easy to get bogged down and lost in your notation; I would try harder were it not for the simple fact that, whatever you may conclude with the series you're dealing with, it still has nothing to do with the original sets involved. How many times must I say it? Sets and series are two different things.

Anyway, as I said, I'm having trouble following your point in 2b. If you don't mind, I'd prefer we start with your example 2, then work our way up to 2b if we have to.

If I insert, then remove, an element from a set, it's gone. This is why, in the original example, the set winds up empty--every element ever inserted into the dish has been removed.

If I add, then subtract, a number from an infinite series, is it "gone"? Not necessarily. Your example 2:

1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 +...

Your analysis of this series:We've just proved that in example 2, the fly ends up at the origin.
I assume you say this because of your Thereom:The set D of moves to the right not compensated for by an equal move in the -X direction (hereafter "antimove") at two hours is empty.
This theorem doesn't say anything about the sum of a series, it deals only with a set of the terms. I see no contradiction with this theorem and the fact that the series sums to log(2).

Here's what can happen in series. Take our favorite:

1 + 1/2 - 1 + 1/3 + 1/4 - 1/2 + 1/5 + 1/6 - 1/3 +...

Rearrange it so the first terms are all positive for a while:

1 + 1/2 + 1/3 + ... + 1/n

I don't know what n is, let's just say it's really huge, like on the order of that bound for Graham's number that Chronos mentioned. Now stick in

-1.

Now let's continue with the positives again, say, for as many terms as we did the first time. Now stick in

-1/2

and continue with a huge string of the positive terms, and so forth.

Now if consider the set of moves and antimoves, as you were doing, all the moves and antimoves cancel each other out. The set of things that don't get cancelled is empty.

That's because, being a set, is doesn't matter at what "rate" (whatever that means) things are being put in or taken out, or anything like there. Think of the set as a static object--either the element is there, or it isn't, nothings "moving around" or going anywhere; it's just a set, sitting there. Everything put in it was taken out, so it's empty.

Conversely, it can be helpful to think of the series, in some sense, as being dynamic, as you were doing with the fly analogy. The rate at which things are coming now does matter, and clearly in the above reordering, the positives are coming at such a fast rate they are "overpowering" the negative terms.

Again, think of the set as just "sitting there"--if you remove everything it ever contained, it's gotta be empty.

And think of the series as being dynamic, the rate at which the terms now (and only now) come becomes important, and even if things cancel each other out in the sense of the sets, the "antimoves' may be spread too thinly to make a difference in the sum.

Cabbage
12-15-2002, 11:42 PM
Oh, forgot this:What if God always writes an integer on each chip, in addition to the dots, and equal to the number of dots (writing the new number, then erasing the old if he adds a dot)? What numbers are written on those chips with infinitely many dots?

Well if he's only writing integers on the dots, I'd say no number is written on the chips at the end--the chip just has the dots on it.

panamajack
12-16-2002, 02:49 AM
Cabbage has already explained (better than I would have) something I realized about case IIIC. I was at first thinking it actually was similar to the OP Case 1, but as he explained, you wouldn't get the integers on the chips. So I apologize for any confusion I created over that.

One other thing about you argument, knock knock :

As long as there are a finite number of chips in the dish, you do not need to worry about which choice you make, since you can make the number of chips in the dish end up infinite by always picking those chips that are in the dish after this move. So, whether the old chip goes back into the dish or the new chip goes in on this move is irrelevant. Since the move n is arbitrary, the choice of chip is irrelevant on any move.

I may be reading this wrong. You seem to be basing the irrelevance of your choice the current move on the fact that you can 'make up for it' in the future, and then concluding that you therefore never have to make up your mind. It sounds like simple procrastination. You actually do have to choose at some point, otherwise the argument doesn't work.

I also gave a slight amount of thought to the variation I brought up involving random choosing of method (which has only slight relevance to this). I think the expected number of chips would be zero now. Given an infinite number of trials, the prob. that chip 1 is in the cup goes to zero, and the same holds for the next highest integer. But then again you need an infinite number of trials for each integer for this to work (which unless I'm mistaken is an uncountably infinite number of trials), so I'm probably wrong again.

NE Texan
12-16-2002, 09:07 AM
Here's a thought...

Cantor had a proof that the real numbers had the same cardinality (a measure of infinity) as the set of all possible subsets (including infinite subsets) of rational numbers. It worked something vaguely like this:

First, you take all rational numbers, and order them. You can do this - it's possible to put them into 1 to 1 correspondence with the natural numbers.

Now given any subset of the rationals, you can construct the following real number. You write it in binary. It starts 0. , then, the next digit is a 1 if the first rational number (in your ordering scheme) is in this subset, and a 0 if not. The next digit is a 1 if the second rational number is in the set, and a 0 if not. An so on...

A couple of examples here:

Let's say your ordering of the rationals begins (0, 1, -1, 2, -2, 1/2, -1/2, 3, ...)

The set {0, 1} is represented as 0.11 (binary), which is 3/4
The set {0,1, 2} is represented as 0.101 (binary), which is 5/8
The set of natural numbers would be an irrational number starting with 0.10100010...

The set of all rationals (which is the largest subset) would be represented by 0.111111.... which is 1.0

The empty set would be represented by 0 (it's 0.00.....)

So, coming back around to this thread (you were wondering about that, weren't you?) - the set from the first page where the smallest numbers were continually taken out would be:

0.000000.....

Which is 0. The empty set. It has no elements. The bowl must be empty.

ZenBeam
12-16-2002, 04:54 PM
Sets and series are two different thingsI've already responded to this.

This theorem doesn't say anything about the sum of a series, it deals only with a set of the terms. I see no contradiction with this theorem and the fact that the series sums to log(2).I'm still thinking about this one.

I'd say no number is written on the chips at the end--the chip just has the dots on it.This argument is why I specifed the new number be written first, then the old number removed. The chips never have no numbers written on them.

Tyrrell McAllister writes
I think that an important change is made when we switch from decimal notation to dot-notation. Decimal notation, by definition, can only represent finite quantities, but dot-notation is much more powerful: it can be used to represent infinite quantities. In fact, dot-notation could be used to represent any infinite cardinal whatsoever. Therefore, this is a very non-trivial change to the parameters of the problem.Interestingly, I don't think anyone has considered what happens if the dot-notation is applied to the OP. In the OP:say god got bored and made a bunch of chips with numbers written on them 1 to infinitytake "numbers written on them" to mean (or be replaced with) "numbers of dots marked on them". Is the answer to the number of chips in the dish at t=2 different? I'd expect the answer not to be different, but since I'm not seeing eye-to-eye with several of you, I don't feel I can assume everyone here agrees with that.

Cabbage
12-16-2002, 05:45 PM
ZenBeam, this seems to me to be the major flaw in your argument: The set of moves and antimoves satisfying conditons (i), (ii), and (iii) do not contribute to the fly's final position (they were either never made, were compensated for, or were compensating a move).
A set is entirely defined by the elements it contains (Axiom of Extensionality (http://mathworld.wolfram.com/AxiomofExtensionality.html)), nothing more.

In particular, say I have two sets, one set P consisting of positive real numbers (just some collection, not necessarily all of them, in fact, let's say it's a countable set, considering what is to follow), the other set N consisting exactly of the negatives of the reals in the previous set. I define a new set:

X = {real numbers x : (x is in P but -x is not in N) or (x is in N but -x is not in P},

then indeed this set X is empty. A set is defined by the elements it contains, and we can easily demonstrate, for each real number x, that x is not in X, hence X is empty.

What happens when we try to form a series, where each term is coming from set P or set N? Let's say the series uses all of the terms from P and N (we can do this, since I stipulated that both are countable).

Clearly we have a situation similar to yours--every "move" is counteracted by an "antimove", and vice versa, by the way I defined P and N.

Can we then conclude the sum is zero (since a given move must always be counteracted by its antimove, and vice versa)? Definitely not. Any number of things can be a problem, but let's focus on rearrangements.

(Now this is where the ordering becomes important).

As I mentioned in my previous post, imagine an incredible number of positive terms leading off the series, followed by a single negative, followed by another slew of positives, again followed by a single negative term, and so on.

Now (and I'm being somewhat vague since the details would depend on the specific series, and I'm trying to explain this intuitively), the positive terms are coming so quickly, some of them may not be "cancelled" by their corresponding negative term for quite a while. In fact, if we look at the partial sums, we will see that at all times, there are positive terms that have yet to be cancelled.

And how is it that we define the sum of an infinite series? That's right, as the limit of the partial sums! And since no partial sum will consist of just "moves" and their corresponding "antimoves" (the negatives are coming to slowly), this property of the partial sums may very well get carried over into the limit. Hence, there's absolutely no reason to expect that the infinite series will sum to zero.

Note, once again, that this is entirely different from the way the sets themselves behave. In that case, it was simply a matter of, "Well, this x is in P, let me look around and see if I can find a corresponding -x in N...Ah, there it is! Well, they cancel, so my set X is still empty", and that's all there is to it--order has no bearing on it.
This argument is why I specifed the new number be written first, then the old number removed. The chips never have no numbers written on them.
Well, they have never have no numbers written on them so long as there are numbers left to write. What happens when we have passed through all the numbers, and there are no numbers left to write?
Interestingly, I don't think anyone has considered what happens if the dot-notation is applied to the OP.
Yeah, I don't think this is any different from the problem in the OP.

I've got a question that I was wondering about last night. For those who claim (in the OP's problem) that the dish is not empty, one thing was never made clear to me. Do you claim that it contains integers? Or do you claim that it magically, somehow, contains "something else"?

If it contains "something else", how can that possibly be so, when, by definition of the problem, God only puts natural numbers in the dish? Did the "something else" just, I don't know, somehow materialize there?

To claim that the dish contains..."something else"...epitomizes the very notion of "hand waving".

Tyrrell McAllister
12-16-2002, 06:49 PM
In response to knock knock's Case II, described in this post (http://boards.straightdope.com/sdmb/showthread.php?postid=2709764#post2709764), I wrote,
God has 1-kg of gold, broken into infinitely many peices, each of which is 0-kg. In fact, if I'm not mistaken, he has uncountably many pieces...
To this, knock knock replied
In this case I believe you are mistaken, possibly because I wasn't quite clear to begin with. On each move, God only breaks one of the pieces of the cylinder in half (and it's always one of the largest remaining pieces of the cylinder). I was trying to keep this problem isomorphic with the original. The number of pieces of cylinder after n moves = number of dots on coin after n moves = number of chips in dish after n moves. The number of pieces increases by 1 each move, so it stays countable.
Sorry to have taken so long to reply to this. I still believe that there are uncountably many coins at 2 hours in this scenario. The reason is that the splitting of the coins over the course of the 2 hours produces an infinite binary tree, and the coins that exist at the 2 hour mark are the terminal nodes of that tree. Infinite binary trees have uncountably many terminal nodes because there is a bijection between the terminal nodes and real numbers written in base 2. In particular, a given terminal node may be described by the sequence of left and right branches that must be traversed to get to that node from the root. So, the node I'd get to by always going left would be 0.00000000...; the node I'd get to by always going right would be 0.11111...; the node I'd get to by going first left, then right twice, and then only left, would be 0.01100000...; and so on.

That this scenario seems isomorphic to the case in the OP so long as everything is finite is another good example of how things can go hay-wire when things become infinite, in the sense that what was irrelevant in the finite case suddenly dramatically affects the outcome.

ZenBeam has been trying to use the reasoning from my proof in this post (http://boards.straightdope.com/sdmb/showthread.php?postid=2703767#post2703767) to derive a contradiction. In his Example 2 (http://boards.straightdope.com/sdmb/showthread.php?postid=2704534#post2704534), he argues that my reasoning leads to one conclusion, while in Example 2b (http://boards.straightdope.com/sdmb/showthread.php?postid=2714362#post2714362), he argues that the same reasoning leads to another contradicting conclusion. Therefore, he concludes, my reasoning is invalid.

It seems to me that there are several problems, in both of these proofs, outside of the portions borrowed from me. Only after both arguments are logically unimpeachable, except for those steps exactly mirrored in my proof, can we conclude that my proof is invalid.

Rather than detail every questionable point now, I will ask Zenbeam to clarify one point at a time, moving on to the next as each difficulty is resolved, until we reach a conclusion.

So, first of all, in ZenBeam's account of Example 2b, he writes,
The set of moves and antimoves satisfying conditons (i), (ii), and (iii) do not contribute to the fly's final position (they were either never made, were compensated for, or were compensating a move). The remaining moves and antimoves satisfying (iv) and (v) can now be easily summed, since they form an absolutely convergent sequence:

1 - 1/6 - 1/20 - 1/42 - ...

The sum converges (absolutely) to log(2). So for example 2b, the fly ends up at log(2).

Here ZenBeam seems to be reasoning as though series had the following property:
Given a series San, if some subset of the terms in San may be canceled with each other (ie, "compensate" for each other), and if the series Sbn that results from removing this subset of terms from San is absolutely convergent, then San = SbnBut this is certainly not true, as may be seen by comparing
1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + . . .
= 1 + 0 + 0 + 0 + . . .
= 1 (absolutely)

and

(1 + -1) + (1 + -1) + (1 + -1) + . . .
= 0 + 0 + 0 + . . .
= 0 (absolutely).

To draw his conclusion, ZenBeam, must show just how my reasoning implies that this method of evaluating series is valid. But this seems highly improbable, because my argument never involved what it even means to perform addition on infinitely many numbers. The entire apperatice of limits, sequences, and complete fields must be defined to do this, all of which are far outside the scope of my proof. So since I made no claims about how series should be evaluated, it is hard to see how I made any wrong claims about how series should be evaluated.

ZenBeam
12-16-2002, 10:02 PM
And how is it that we define the sum of an infinite series? That's right, as the limit of the partial sums! And since no partial sum will consist of just "moves" and their corresponding "antimoves" (the negatives are coming to slowly), this property of the partial sums may very well get carried over into the limit. Hence, there's absolutely no reason to expect that the infinite series will sum to zero.

Note, once again, that this is entirely different from the way the sets themselves behave.The OP is not asking what is the set of chips in the dish. The OP is asking how many chips are in the dish. This is not a set, this is a series. I have been trying to use your set arguments to show a contradiction in a similar problem, and the response has been that I can't use sets for that. Why should any of us believe you can use sets to obtain the answer to the OP? To use your own words, there's absolutely no reason to expect that the infinite series will sum to zero.

Tyrrell McAllister writes:
Here ZenBeam seems to be reasoning as though series had the following property:

Given a series San, if some subset of the terms in San may be canceled with each other (ie, "compensate" for each other), and if the series Sbn that results from removing this subset of terms from San is absolutely convergent, then San = Sbn

But this is certainly not true, as may be seen by comparingObviously this is untrue, yet that is what your argument is doing when showing that the number of chips is zero. You say you made no claims about how the series should be evaluated, but you did. You evaluated the series by canceling chips added with chips removed, then summing what was left (an empty set). Again, if this does not work on the fly, why should we have any confidence it works on the chips?

If it contains "something else", how can that possibly be so, when, by definition of the problem, God only puts natural numbers in the dish?If it contains zero chips, how can that possibly be so, when, by definition of the problem, God only increases the number of chips at every step? I've addressed your question three times previously in the thread so I'll let someone else answer.

This theorem doesn't say anything about the sum of a series, it deals only with a set of the terms. I see no contradiction with this theorem and the fact that the series sums to log(2).Fine. By this same reasoning there's no contradiction between the empty set of chips and an infinite number of them. What's on the chips? Who cares? Maybe they're blank. In response to this:
What if God always writes an integer on each chip, in addition to the dots, and equal to the number of dots (writing the new number, then erasing the old if he adds a dot)? What numbers are written on those chips with infinitely many dots?you were willing to accept blank chips:
Well if he's only writing integers on the dots, I'd say no number is written on the chips at the end--the chip just has the dots on it.So you shouldn't have any difficulty with that.

OK, this whole post is kind of snarky. It's late, and I'm not gonna spend the time to de-snark it. I'll just apologize in advance.

Cabbage
12-16-2002, 11:03 PM
ZenBeam, so the original problem isn't a set, after all?

We've got a dish, which we can think of as the set. We've got the numbered chips, which we can think of as elements that either will or won't be in the set at the end. And to finish it off, we have a cardinality function, which tells us how many elements are in any given set. This is all we need to answer the original question.

A couple of questions:
1. Do you agree that modeling it as a set, the set is empty at the end?

2. How would you justify that it's not a set? Would you justify that it can't be a set simply because the conclusions are counterintuitive and don't meet your expectations? (I would certainly hope for a better justification than that).

knock knock
12-17-2002, 09:35 AM
Originally posted by Tyrrell McAllister
So, first of all, in ZenBeam's account of Example 2b, he writes,
"The set of moves and antimoves satisfying conditons (i), (ii), and (iii) do not contribute to the fly's final position (they were either never made, were compensated for, or were compensating a move). The remaining moves and antimoves satisfying (iv) and (v) can now be easily summed, since they form an absolutely convergent sequence:

1 - 1/6 - 1/20 - 1/42 - ...

The sum converges (absolutely) to log(2). So for example 2b, the fly ends up at log(2)."

Here ZenBeam seems to be reasoning as though series had the following property:
Given a series San, if some subset of the terms in San may be canceled with each other (ie, "compensate" for each other), and if the series Sbn that results from removing this subset of terms from San is absolutely convergent, then San = SbnBut this is certainly not true, as may be seen by comparing
1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + . . .
= 1 + 0 + 0 + 0 + . . .
= 1 (absolutely)

and

(1 + -1) + (1 + -1) + (1 + -1) + . . .
= 0 + 0 + 0 + . . .
= 0 (absolutely).

To draw his conclusion, ZenBeam, must show just how my reasoning implies that this method of evaluating series is valid. ZenBeam's series: log(2) = 1 + (1/2 - 1) + 1/3 + (1/4 - 1/2) + 1/5 + (1/6 - 1/3) + ...

The problem you've shown with grouping terms only holds for series whose terms do not approach 0. Informally, the definition for the sum of a series is that a series sums to L if you can get as close to L as you want just by going far enough out in the series (i.e. adding up terms in a partial sum) and you stay that close once you go farther out). We know that the series for log2 converges, and that, if you've added up 3n terms in ZenBeam's series you'll have the same thing as if you'd added up 2n terms in the series for log2. So I can get as close as I want to log2 by going out far enough in that series, so, if I stick to looking at ZenBeam's series after 3n terms (i.e. in chunks of 3) I can get as close as I want to log2. And, since all the terms are getting small, I won't get too far from log2 on the 3n+1 and 3n+2 terms. So the series converges to log2.

Maybe you can argue against using series at all, or using any definition of convergence, but don't worry about ZenBeam's way of adding up the series.

Cabbage
12-17-2002, 03:29 PM
knock knock:ZenBeam's series: log(2) = 1 + (1/2 - 1) + 1/3 + (1/4 - 1/2) + 1/5 + (1/6 - 1/3) + ...

The problem you've shown with grouping terms only holds for series whose terms do not approach 0.
I thought I had killed this one already. ZenBeam's argument was that since every positive term is eventually cancelled by a negative term (and vice versa), the series sums to zero. This simply isn't true. That's a rearrangement, and I've already adressed the problem with doing that, in my first post on this page. And my argument does specifically apply to ZenBeam's series above. In fact, I can rearrange that series so that it doesn't converge at all, so what's the point?

ZenBeam
12-17-2002, 04:21 PM
Do you agree that modeling it as a set, the set is empty at the end?No. I don't believe your argument is valid.

How would you justify that it's not a set? Would you justify that it can't be a set simply because the conclusions are counterintuitive and don't meet your expectations?The answer the OP is looking for is not a set, and I shouldn't have said is was a series. It's a value. The problem expressed as a series would gives us the value if only the series were convergent, but it is divergent. You try to get around this by expressing the problem in terms of sets. You believe that by using sets, you aren't rearranging terms, and can therefore do this (I hope I'm not putting words in your mouth here, but I really don't think you'll disagree with that statment). I believe what you are doing is as invalid as rearranging the terms of a series, and is likely equivalent to doing so. I base this on my examples 2 and 2B, where I express the problems as sets and run into the same errors as I would run into rearranging terms in a non-absolutely converging series.

I'm not certain precisely the issue causing the difficulty in expressing the problem in terms of sets. Regardless, the contradictory solutions to examples 2 and 2B are sufficient for me.

This just occured to me as I was writing the above paragraph. I'll confess I haven't given it much thought, but I'll throw it out here anyway. Perhaps the issue with using sets is that the sequence of sets of chips obtained at each step does not converge to your final set (and in fact does not converge at all). You've requested I show continuity between finite and infinite series. Can you (or anybody) show continuity between the sequence of finite sets at each step, and the final, infinite (or empty) set?

ZenBeam's argument was that since every positive term is eventually cancelled by a negative term (and vice versa), the series sums to zero.The point of my argument is that this is what your and Tyrrell McAllister's arguments using sets are doing. Let me be clear here: I understand full well that this is not allowed in summing series which are not absolutely convergent. I am following the set arguments attempting to show the dish is empty when I do this. As I said above, I believe what you are doing is as invalid as rearranging the terms of a series, and is likely equivalent to doing so.

Cabbage
12-17-2002, 05:25 PM
This just occured to me as I was writing the above paragraph. I'll confess I haven't given it much thought, but I'll throw it out here anyway. Perhaps the issue with using sets is that the sequence of sets of chips obtained at each step does not converge to your final set (and in fact does not converge at all). You've requested I show continuity between finite and infinite series. Can you (or anybody) show continuity between the sequence of finite sets at each step, and the final, infinite (or empty) set?

My point all along has been, why would it have to converge or be continuous in the first place. All I've done is merely investigate the sets and let them speak for themselves.

There's an analogous thought experiment I've been tossing in my head for a few days now, which may be worth bringing up. Somehow, to me, this experiment makes the conclusion somewhat more intuitive, and I'm hoping that it may do the same for you.

We have a village, like any other village, except this village has a (countably) infinite population. The people of this village are numbered, 1,2,3,4,.... No person has two numbers, no number is on two different people, everybody gets a number, every number is on somebody, blah blah blah. What I'm saying is there is a 1-1 correspondence between the numbers and the people.

Anyway, the people of this village live in houses, like most villagers do. And this village has curfew at midnight--everyone must be in their house at midnight. Let's even say that every night, exactly at midnight, a dragon swoops through the town, breathing fire over the entire village, killing everyone in sight. Which, of course, would explain why their houses are fireproof).

On one particular night, at 10PM, persons 1 and 2 go out to enjoy the weather, or whatever. At 11PM, persons 3 and 4 also go outside, while person 1 goes back home. At 11:30, 5 and 6 go out, while 2 goes home. And so on...I think we all know the setup by now.

As it gets closer and closer to midnight, more and more people are coming outside. But watch any particular person, say the person numbered N. You will notice that 1/2N-1 hours before midnight, person N heads home. The same is true for everyone else.

I can even imagine a census taker, having all the people line up in the morning. As he goes down the line, he asks each person what time he came home last night. Each person tells him "I made it home before midnight and was safe from the dragon". The census taker checks off every one on his list.

"All present and accounted for".

How many people broke curfew, and were out at midnight? Was anyone killed by the dragon? And...dare I ask it again...who, if anyone, was killed?

Tyrrell McAllister
12-17-2002, 05:50 PM
Originally posted by ZenBeam
The OP is not asking what is the set of chips in the dish. The OP is asking how many chips are in the dish. This is not a set, this is a series.
As Cabbage has pointed out, the OP is asking for the cardinality of a set. Set cardinalities may not, in general, be found using series. Before using series to find this cardinality, you must first justify that it is valid to do so in this particular case. At the very least, this will require making some kind of continuity argument about how the cardinality of the set changes as a function of time.

I think that you are confusing two distinct usages of the word "add". On the one hand, we say that we are "adding something up" when we are counting the elements of (i.e.finding the cardinality of) a set. On the other hand, we say that we are "adding" when we perform the binary operation of addition on some elements of R, the set of real numbers. These two notions of "adding" are different. The first case may be applied to sets of arbitrary size, and the "adding" here is done using bijections and the other tools of set theory. In the second case, the machinary of limits and so forth must be brought into play to perform this kind of adding on a nonfinite number of elements of R.

Obviously this [claim about how series may be evaluated] is untrue, yet that is what your argument is doing when showing that the number of chips is zero. You say you made no claims about how the series should be evaluated, but you did. You evaluated the series by canceling chips added with chips removed, then summing what was left (an empty set).(bolding added)
This is an example of the confusion I referred to above. I summed no series; I evaluated cardinalities.

I will try to give another example of how these two notions of "adding" are different. Consider the subset S of R below:
{ -1/1 , 1/2 , -1/3, 1/4, -1/5 , . . . }If I "add up" the elements in this set using the first meaning, I will get a certain cardinality (aleph-0), and I will get this answer no matter what order I count them in. But if I "add up" these element by evaluating a series whose terms are these elements, I can get any answer whatsoever, depending on the order of the terms in the series. So, again, "adding up" can mean two very different things.

You wrote
If it [the dish in the OP] contains zero chips, how can that possibly be so, when, by definition of the problem, God only increases the number of chips at every step?
This is just one of the bizarre phenomena that can occur when things get infinite. Consider my set S above. Suppose I removed all elements of this set, one after the other, at decreasing intervals so that I've removed every element after two hours. Then no matter what order I do this in, the set that results at 2 hours will be empty. "But," you might say, "how can that possibly be so, when, by definition of the problem, you have infinitely many elements in S at each step?" What can I say? It may be unintuitive, but it is the logical conclusion.

knock knock wrote
The problem you've shown with grouping terms only holds for series whose terms do not approach 0.
I still don't think the principle works in general, for the reasons Cabbage has been pointing out. For example, a series converging to 1 may be constucted whose terms are elements in the set
S = {1/n: n in N} union {-1/n: n in N}(Since the series converges, the terms tend to zero, and so satisfy your additional condition.) Now "interleave" this series with some absolutely convergent series, like 0+0+0+... to construct the series San. Though the terms in S cancel (in the sense that for each 1/n, there is a compensating -1/n), the value of San is different from the value of 0+0+0+... that results from the removal of the terms in S from San.

Tyrrell McAllister
12-17-2002, 05:57 PM
I wrote
For example, a series converging to 1 may be constucted whose terms are elements in the set
S = {1/n: n in N} union {-1/n: n in N}
I should have added that I mean for all the elements in S to be terms in the series. That is, every term in the series is an element of S, and every element in S is a term in the series.

knock knock
12-17-2002, 10:29 PM
Sorry, Tyrrell McAllister, I misunderstood what you were saying about ZenBeam's post. I didn't read your post closely enough, so I thought you were objecting to him grouping his terms by putting parentheses around them. You were really talking about rearrangements, and in that case you're correct that you can't rearrange an infinite series and say that you still have the same series, because it can converge to something else unless the series is absolutley convergent. I'm not ready to say what implications (if any) this has on ZenBeam's argument.

Originally posted by Tyrrell McAllister
Sorry to have taken so long to reply to this. I still believe that there are uncountably many coins at 2 hours in this scenario. The reason is that the splitting of the coins over the course of the 2 hours produces an infinite binary tree, and the coins that exist at the 2 hour mark are the terminal nodes of that tree. Infinite binary trees have uncountably many terminal nodes because there is a bijection between the terminal nodes and real numbers written in base 2. In particular, a given terminal node may be described by the sequence of left and right branches that must be traversed to get to that node from the root. So, the node I'd get to by always going left would be 0.00000000...; the node I'd get to by always going right would be 0.11111...; the node I'd get to by going first left, then right twice, and then only left, would be 0.01100000...; and so on.

That this scenario seems isomorphic to the case in the OP so long as everything is finite is another good example of how things can go hay-wire when things become infinite, in the sense that what was irrelevant in the finite case suddenly dramatically affects the outcome.
I'm not so sure about this. For one, there seems to be a perfect map between the coins and the pieces of the cylinder. On one move, you break a piece of cylinder in two, and if you want, you could call one piece the "old piece" and the other the "new piece." With the coins, you can stack the coins, and on one move you can place a new coin next to one of the old coins, and there is a pattern to where you place the new coin that can mimic the cylinder problem. If you wanted to, you could even keep shrinking the coins, so that the two problems would look identical.

I'm not sure why you're mapping the cylinder pieces to base-2 "decimals" like .01, .1101, & .001001 insteading of mapping them to the corresponding base-2 integers, like 10, 1011, & 100100. Whichever map you choose (and my inclination is that the map to the integers is more fitting), it looks like you should be able to use the same map for either case.

All of this is irrelevant to the OP, of course.

I like your dragon example, Cabbage. I'll have to think about it some more.

Tyrrell McAllister
12-18-2002, 04:44 AM
Originally posted by knock knock
I'm not sure why you're mapping the cylinder pieces to base-2 "decimals" like .01, .1101, & .001001 insteading of mapping them to the corresponding base-2 integers, like 10, 1011, & 100100. Whichever map you choose (and my inclination is that the map to the integers is more fitting), it looks like you should be able to use the same map for either case.

All of this is irrelevant to the OP, of course.

Let me try to explain my thinking in more detail. Consider a binary tree:

/\
/ \
/ \
/ \
/ \
/ \
/ \
/ \
/\ /\
/ \ / \
/ \ / \
/ \ / \
/\ /\ /\ /\
/ \ / \ / \ / \
/\ /\ /\ /\ /\ /\ /\ /\
:: :: :: :: :: :: :: ::
__________________________________

In this tree, the branches at each level are half the length those on the preceding level, so the tree ends at the horizontal line at the bottom.

The tree has countably many branch points (nodes) because you can enumerate them by labeling the node on the first level 1; the nodes on the second level 2 and 3; the nodes on the third level 4, 5, 6, and 7; and so on.

But consider the set of all paths down the tree. This set is uncountable, because I can define a bijection f between these paths and the binary numbers between zero and one as follows:
f(P) = 0.e1e2e3e4..., where P is a path and, for each i = 1, 2, 3, . . . , I define ei = 0 (resp. 1) if path P takes the left (resp. right) branch at the ith node on the path. (It is because each path passes through infinitely many nodes that I need to use the digits on the right of the decimal point, rather than on the left, to represent them).

The horizontal line at the bottom is supposed to be all the "limit points" you can get to by traveling one of these paths down the tree (I called these points on the line "terminal nodes" in my other post). There is exactly one of these points for each path, so there are uncountably many of these points. The horizontal line is also like the cylinder of gold in you scenario. Each terminal node is like a coin in the cylinder at the 2 hour mark.

Now, the binary tree in your scenario is arranged a little differently if we want the vertical axis to be time. Here, there is only one branch point at each level before the horizontal line. So we have something more like:

/\ 0 hours
/ \
/ \ |
/ \ |
/ \ |
/ \ |
/\ \ | Time
/ \ \ |
/ \ \ |
/ \ /\ |
/\ \ / \ |
/ \ /\ / \ |
/ \ / \ /\ \ \ /
/ \ / \ / \ /\ v
: : : : : : ::
________________________________ 2 hours

But my function f still works, and so still establishes the uncountability of the coins at 2 hours.

ZenBeam
12-18-2002, 01:18 PM
There's an analogous thought experiment I've been tossing in my head for a few days now [...] I can even imagine a census taker, having all the people line up in the morning. As he goes down the line, he asks each person what time he came home last night. Each person tells him "I made it home before midnight and was safe from the dragon". The census taker checks off every one on his list.
"All present and accounted for".If Science Fiction has taught me anything1, it's that in the next instant, the census taker turns to look at a large number of bodies charred beyond recognition, stretching to the horizon, saying "...so who are all those people." [Cue the spooky music, fade to black.]

1. Here's a free straight line.

William_Ashbless
12-18-2002, 02:05 PM
Um, at the risk of being obtuse and dim, it seems to me that the exact number of coins at the two hour mark is akin to some sort of critical point that cannot easily be analysed, like 1/x at x=0. Does it much matter how many coins there are at exactly the two hour point? Isn't it of much more interest to decide on the state of things immediately before or immediately after the two hour point?

Cabbage
12-18-2002, 03:51 PM
Isn't it of much more interest to decide on the state of things immediately before or immediately after the two hour point?
Not if we're specifically interested in what happens at the two hour mark (or after, for that matter, since nothing has been said to happen after the first two hours, the dish will still be empty after two or more hours, but not before).

ZenBeam, proof by science fiction? Gimme a break. :rolleyes:

By definition of the problem, there are no other people. Are you claiming this process of people leaving and entering their houses is some kind of...(waving hands)...voodoo ritual causing other beings to manifest themselves in the village? Are you even considering my arguments, or are you just convinced that you are right and won't listen to reason? Despite what you may think, I've already explained what's wrong with your "refutation" involving series. Twice.

Here's an idea, pretend these fictional people you bring up are, well, fictional. They don't exist. All we have in the village are the numbered people, period. What do you say now?

ZenBeam
12-19-2002, 09:54 AM
Here's an idea, pretend these fictional people you bring up are, well, fictional. They don't exist. All we have in the village are the numbered people, period. What do you say now?The numbered people are fictional too. It sounds like you want me to assume the numbered people are real, and no science fiction allowed. This problem is ill-posed. You can't have an infinite number of real people. The observable universe isn't big enough. Once you go outside the observable universe, you're into fiction. There are many other problems with trying to set up this problem in a physically real way.

Mathematically, your village is equivalent to the OP, and I've addressed this issue three times previously, and this is now the second time I've said that I adressed it three times previously. You want a fourth? If the number of chips in the dish, or people outside their houses, is undefined, there's no contradiction in not being able to name one.

Here's another setup exploring the name them issue, and I hesitate to bring it up, since this thread is full of them, and quite frankly this thread is wearing me out, but this one is pretty simple. Assume God has two dishes, A and B, and two chips, 0 and 1. At t=0, God puts chip 0 in dish A and chip 1 in dish B. At t=1 hour, He switches them, removing chip 0 from dish A, and placing it in dish B (to be precise, I'll specify chip 0 is in dish A for t<1, and in dish B for t>=1), and simultaneously, chip 1 is likewise moved to dish A. At t=90 minutes, he switches them again, again at t=105 minutes, and so forth. At t=2, how many chips are in the union of the two dishes? This has to be two, right? How many chips are in each dish? Doesn't the answer have to be 1? (I guess it could possibly be undefined.) If the answer is 1, at t=2, which chip is in which dish? This has got to be undefined.

While we're here, what if I try to determine how many chips are in dish A? For every time chip 0 is put in dish A, I can tell you when it was removed, just like in the OP. Same for each time chip 1 is placed in dish A. Does that mean dish A is empty? The same argument applies to dish B. But the union of dishes A and B has two chips, so where are they?

You can describe this problem with two villagers and two huts, but mathematically it's the same problem, with the same answers (or lack thereof). Physically, you don't have the problem of an infinite number of people, but you do run into infinite velocities as t-->2, and quantum mechanically there'd be issues also.

Cabbage
12-19-2002, 12:41 PM
The numbered people are fictional too. It sounds like you want me to assume the numbered people are real, and no science fiction allowed. This problem is ill-posed. You can't have an infinite number of real people. The observable universe isn't big
I meant "fictional" in the context of the problem, of course. In the context of Lord of the Rings, Gandalf is real, but Darth Vader is fiction. The people/chips and so on are merely ways to help think of this purely mathematical problem, so who cares if I can have an infinite number of people, either. I certainly have an infinite number of integers, and I can phrase the problem in terms of those. Or are you saying that all abstract thought is meaningless, since it's not real?
If the number of chips in the dish, or people outside their houses, is undefined, there's no contradiction in not being able to name one.
But it's not undefined, every set has a cardinality. The cardinality of the collection of people in the village is omega (or aleph-naught, if you prefer.). The cardinality of the chips involved in the problem is, again, omega. The cardinality of any countably infinite set is omega. And since, in each case, our "universal" set of chips/people is countable, we certainly know that the number of chips/people left has either finite cardinality or omega cardinality.

I would agree that your example is not well defined. Since the chips switch at every step, it's not defined which is where at the end.

This is a significant difference from the OP, where, once a chip has left, it never returns to the dish.

This really is nothing more than a simple application of mathematical induction (http://mathworld.wolfram.com/InductionPrinciple.html). I can demonstrate that chip one is not in the dish, I can demonstrate that if chip n is not in the dish, then neither is chip n+1 (since it's the very next to be taken out). Therefore, none of the chips numbered 1,2,3,... are in the dish. Finally, since there are no other chips we can conclude the dish is empty.

I'm getting a little tired of dealing with this thread as well, and if I can't convince you of the above induction argument, there's probably not much point in either of us prolonging it.

knock knock
12-19-2002, 03:41 PM
Originally posted by Cabbage
This really is nothing more than a simple application of mathematical induction (http://mathworld.wolfram.com/InductionPrinciple.html). I can demonstrate that chip one is not in the dish, I can demonstrate that if chip n is not in the dish, then neither is chip n+1 (since it's the very next to be taken out). Therefore, none of the chips numbered 1,2,3,... are in the dish. Finally, since there are no other chips we can conclude the dish is empty.This is not a simple case of mathematical induction, because induction doesn't always work so well when you have an infinite set. I argued by induction against the no chips left conclusion, on examples IIIC/IIID, where on each move you put in a chip, take out a chip, renumber the chip that you took out so that it has the same number as the other chip you're supposed to put in, and then choose whether to put in the chip that was just in or the other chip. The conclusion from the "no chips left" crowd was that, if you always put the same chip back, you end up with infinitely many chips, but, if you always put the other chip in, you end up with 0. So the final result depends on your choices. But if you put in the new chip on the 1st move, that doesn't change anything - there's still 2 chips. If you put in the new chip on the nth move, that doesn't change anything - there's still n+1 chips. For any n, regardless of what you do on the first n moves, it's still possible to end up with infinitely many chips. Your nth move does not matter. So now you can just do induction, or let n go to infinity, or say that if it holds for any n, it holds for all n. And so there are infinitely many chips no matter what you choose on each move. So there are infinitely many chips in the situation in the OP.

I don't know if there's a way to reconcile this argument (and the other ones for infinitely many chips) with your arguments for 0 chips. I'm suspecting this situation is more like the lightbulb problem or the 2-chip problem that ZenBeam just gave than I thought it was, and that the answer is undefined.

I'm going to be traveling for a little while, so I don't know if I'll be able to keep up with this thread very well. Unless anyone has some new insight, it's fine with me if this thread dies.

Chronos
12-19-2002, 04:27 PM
I still like the idea of numbering by dots.

The original problem, expressed this way, goes something like this:
God puts a chip in the bowl with one dot.
God then, all at the same time, puts two chips in, with two and three dots respectively, and takes out the one with one dot.
He then puts in chips with four and five dots, and takes out the one with two.
He puts in chips with six and seven dots, and takes out the one with three.
Et cetera.

Now, are we agreed that this is equivalent to the original problem?

On the first move, God puts in a chip with one dot.
On the second and following moves, he takes out the lowest chip, doubles the amount of dots on it, and puts it back in, along with another chip with one more dot on it.

In this case, what He ends up with is a countably infinite number of chips in the bowl, each of which has a countably infinite number of dots on its surface. There is no integer in the bowl, but there are still chips with symbols on them. It's just that none of the symbols is an integer any more.

In what way, if any, does this differ from the original problem?

Cabbage
12-19-2002, 04:40 PM
This is not a simple case of mathematical induction, because induction doesn't always work so well when you have an infinite set.
Huh? I don't follow you here. The whole point of mathematical induction is to deal with infinite sets. Anybody can deal with finite sets, just take each case individually...you never need induction for that. Induction is specifically for dealing with infinite sets.

And I thought I already mentioned the problem in example IIIC/D; we don't have a direct correspondence between the numbers and the chips. In the original example, you could point at each chip and say, "That's chip 1, that's chip 2,...,that's chip 425463,...", and therefore, you could talk about the problem in terms of the chips, or in terms of the numbers. That's not true of your example; there, you can talk about the chips, or you can talk about the numbers, but there's no longer any guarantee of them behaving the same way, which is why in one case you have infinitely many chips in the dish, another case you have zero chips in the dish. And note that in both cases IIIC and IIID, you have zero integers in the dish, which has been my whole point all along.

Anyway, changing the subject, although I am kind of getting worn out by this never ending debate, I have enjoyed reading and thinking about everyone's contributions. I tend to avoid Great Debates, and often debates in general, usually, because I always worry that someone may take it personally and be personally offended by something I say. I hope that hasn't happened here. I'm glad to have had the discussion, and I hope ya'll are to, but, yeah, I'm about ready to let it drop if everyone else is up for that. And I wish ya'll a Merry Christmas and Happy New Year! :D

panamajack
12-19-2002, 05:38 PM
Okay, I'll give it a chance with another more formal presentation of the problem :

Given B0 = { 0 } (the null set, meaning the bowl starts empty), define

Bi = Bi' - min(Bi')

where min (A) = smallest element of A, and

Bi' = { Bi-1, 2i-1, 2i }

If you do not agree that that is a correct statement of the problem, explain why you feel it is inaccurate.

Consider this :

Proposition-
Pj : For any integer s < j, s is not a member of Bj.

Proof by induction-

B1 = { 2 }. Clearly P1 is true.

Pk => Pk+1?

Pk+1 : For any integer s < k+1, s is not a member of Bk+1

From the set-up, Bk+1' = { Bk, 2k+1, 2k+2 }

If Pk is true, min(Bk) is >= k.
Since 2k+1 and 2k+2 > k, min(Bk+1') >= k. In either case of (>=), there are no integers < k+1 in Bk+1.

Thus Pk => Pk+1, and the proposition is true by induction.

Now, it is possible to show that the number of chips in the bowl increases without bound as i->inf and we approach t=2. However, it's also been shown (several times now) that once we have completed the infinite task, there can be no integers in the bowl.

The fact that removing a countably infinite number of members from a countably infinite set will or won't change the set's cardinality depending on which members you remove may seem strange (and clearly it's not well-understood). It would have been nice to be able to explain it, but like the rest, I'm not sure if it's going to work.

I will agree and say that this is one of the best threads in a long while and I've enjoyed it immensely.

panamajack
12-20-2002, 12:39 AM
Allow me to clarify that last post. First, I had changed my indices, so Pk actually works for s <= k. Replace the appropriate statements with :

If Pk is true, min(Bk) is >= k+1.
Since 2k+1 and 2k+2 > k+1, min(Bk+1') >= k+1. In either case of (>=), there are no integers < k+2 in Bk+1.

Additionally, It is not the intention of that proof to show that the bowl is empty at the end (i.e. for t = [2, oo ) ), because it doesn't show that. Those proofs have already been given by Cabbage, Tyrell McAllister, et al. I just thought that by also observing similar behavior on the other side (leading up to 2), it might cause some light to dawn.

My main intention was to give the formulation I did. Hopefully this will also explain why the problem can't be modeled as a series. A sequence of sets, maybe. BUT we are only concerned with a single set - the terminal set after the operations are completed.

ZenBeam
12-20-2002, 12:25 PM
I agree this has been a fun thread. I haven't always enjoyed it, but I have on the whole.

I'm ready to let it lie too, but I think Chronos's last post needs to be addressed: numbering with knock knock's dots exposes a crucial element. For visuallization purposes, put all the dots in a line on the chip. 1 is represented by a dot at 1, 2 by dots at 1 and 1/2, 3 by dots at 1, 1/2, 1/3, and so forth. The symbol for infinity naturally falls out of this: dots at 1/n for all positive n. As Chronos points out, this symbol isn't an integer. The only difference between Chronos's first example and the OP seems to be the existance of a symbol for infinity which falls naturally from the notation for the integers. This doesn't seem to be a sufficient difference for these two cases to have different answers.

Many of the proofs here show there are no chips with integers on them left in the dish at t=2. This doesn't rule out chips with the symbol for infinity, which therefore must be ruled out some other way. Certainly, there are no such chips for t<2. Has anyone shown there must be no such chips for t = 2? It would seem this can't be said to be obvious, certainly not when using the dot notation, since the numbers of dots on the chips in the dish are approaching infinity. The simplest argument, "no chips ever have an infinite number of dots for t<2, therefore there aren't any at t=2" would seem to be equally as valid or invalid as saying "the number of chips is > 0 for t<2, therefore the number is > 0 for t=2". I don't recall seeing anything more rigorous than that. Also, that argument would apply equally to Chronos's second example above, but I think everyone's going to agree there are chips with infinitely many dots at the end in that case.